answer：Okay, so I have this function f(x) = (-2^x + b)/(2^x + a), and it's given that it's an odd function defined on all real numbers. I need to find the values of a and b. Then, in part II, I have to figure out the range of k such that for any real number t, the inequality f(t² - 2t) + f(2t² - k) < 0 always holds. Hmm, let me start with part I.First, I remember that an odd function satisfies f(-x) = -f(x) for all x. Also, since the function is defined on all real numbers, it should be defined at x=0. So maybe I can use that to find some information about a and b.Let me compute f(0). Plugging x=0 into f(x), I get f(0) = (-2^0 + b)/(2^0 + a) = (-1 + b)/(1 + a). But since f is odd, f(-0) = -f(0), but f(0) = -f(0) implies that f(0) must be zero. So, (-1 + b)/(1 + a) = 0. That means the numerator must be zero, so -1 + b = 0, which gives b = 1.Alright, so b is 1. Now, the function becomes f(x) = (-2^x + 1)/(2^x + a). Now, I need to use the property that f(-x) = -f(x). Let me compute f(-x).f(-x) = (-2^{-x} + 1)/(2^{-x} + a). Hmm, 2^{-x} is 1/(2^x), so let me rewrite that:f(-x) = (-1/(2^x) + 1)/(1/(2^x) + a). To simplify this, I can multiply numerator and denominator by 2^x to eliminate the fractions:Numerator: (-1 + 2^x)Denominator: 1 + a*2^xSo f(-x) = (2^x - 1)/(a*2^x + 1). Now, according to the odd function property, this should be equal to -f(x). Let me compute -f(x):-f(x) = -[(-2^x + 1)/(2^x + a)] = (2^x - 1)/(2^x + a). So, f(-x) = (2^x - 1)/(a*2^x + 1) must equal (2^x - 1)/(2^x + a). Therefore, the denominators must be equal:a*2^x + 1 = 2^x + a.Let me rearrange this equation:a*2^x - 2^x + 1 - a = 0( a - 1 )2^x + (1 - a) = 0Hmm, this equation must hold for all x in R. The only way this can be true for all x is if the coefficients of 2^x and the constant term are both zero. So:a - 1 = 0 => a = 1and1 - a = 0 => a = 1Perfect, so both conditions give a = 1. So, a is 1 and b is 1. That solves part I.Now, moving on to part II. The inequality is f(t² - 2t) + f(2t² - k) < 0 for any t in R. I need to find the range of k such that this holds.First, let me recall that f(x) is an odd function, so f(-x) = -f(x). Also, since f is odd, if I know something about its behavior, maybe it's increasing or decreasing, that could help.Let me first analyze f(x). Since f(x) = (1 - 2^x)/(1 + 2^x). Let me see if I can simplify this or find its properties.Let me write f(x) as:f(x) = (1 - 2^x)/(1 + 2^x) = [ (1 + 2^x) - 2*2^x ] / (1 + 2^x) = 1 - (2*2^x)/(1 + 2^x) = 1 - 2*(2^x)/(1 + 2^x)Alternatively, maybe I can write it as:f(x) = (1 - 2^x)/(1 + 2^x) = (2^0 - 2^x)/(2^0 + 2^x). Hmm, not sure if that helps.Alternatively, let me compute f(x) + f(-x):f(x) + f(-x) = (1 - 2^x)/(1 + 2^x) + (1 - 2^{-x})/(1 + 2^{-x})But since f is odd, f(-x) = -f(x), so f(x) + f(-x) = 0, which is consistent.But maybe I can find if f is increasing or decreasing. Let me compute its derivative to check.f(x) = (1 - 2^x)/(1 + 2^x). Let me denote u = 2^x, so f(x) = (1 - u)/(1 + u). Then, df/du = [ -1*(1 + u) - (1 - u)*1 ] / (1 + u)^2 = [ - (1 + u) - (1 - u) ] / (1 + u)^2 = [ -1 - u -1 + u ] / (1 + u)^2 = (-2)/(1 + u)^2.Since u = 2^x > 0 for all x, the derivative df/du is negative. Therefore, f(x) is a decreasing function of u, and since u = 2^x is increasing in x, f(x) is decreasing in x.So, f is strictly decreasing on R. That's useful.Also, since f is odd, f(0) = 0, and as x approaches infinity, 2^x approaches infinity, so f(x) approaches ( -2^x ) / 2^x = -1. Similarly, as x approaches negative infinity, 2^x approaches 0, so f(x) approaches (1 - 0)/(1 + 0) = 1.So, f(x) is a strictly decreasing function from 1 to -1 as x goes from -infty to +infty.Now, going back to the inequality: f(t² - 2t) + f(2t² - k) < 0.Since f is odd, f(-y) = -f(y). So, maybe I can rewrite the inequality as f(t² - 2t) < -f(2t² - k) = f(k - 2t²).So, f(t² - 2t) < f(k - 2t²).But since f is strictly decreasing, the inequality f(a) < f(b) implies that a > b.Therefore, t² - 2t > k - 2t².Let me write that down:t² - 2t > k - 2t²Bring all terms to one side:t² - 2t + 2t² - k > 0Combine like terms:3t² - 2t - k > 0So, the inequality 3t² - 2t - k > 0 must hold for all t in R.Wait, but the original inequality is f(t² - 2t) + f(2t² - k) < 0, which we transformed into 3t² - 2t - k > 0. So, for all t, 3t² - 2t - k > 0.But that's a quadratic in t. For this quadratic to be positive for all t, the quadratic must be always positive, meaning it never touches or crosses the t-axis. So, the discriminant must be negative.Recall that for a quadratic ax² + bx + c, the discriminant is D = b² - 4ac. If D < 0, the quadratic is always positive (if a > 0) or always negative (if a < 0).In our case, the quadratic is 3t² - 2t - k. Coefficient a = 3 > 0, so if D < 0, the quadratic is always positive.Compute D:D = (-2)^2 - 4*3*(-k) = 4 + 12k.We need D < 0:4 + 12k < 0 => 12k < -4 => k < -4/12 => k < -1/3.Therefore, the range of k is all real numbers less than -1/3.Wait, let me double-check my steps.1. I used the fact that f is odd, so f(2t² - k) = -f(k - 2t²). That seems correct.2. Then, since f is strictly decreasing, f(a) < f(b) implies a > b. That's correct because if f is decreasing, larger inputs give smaller outputs.3. So, t² - 2t > k - 2t². Then, moving terms around: 3t² - 2t - k > 0.4. Then, for this quadratic to be positive for all t, discriminant must be negative. So, D = 4 + 12k < 0 => k < -1/3.Yes, that seems consistent.But wait, let me make sure I didn't make a mistake in the transformation.Starting from f(t² - 2t) + f(2t² - k) < 0.Since f is odd, f(2t² - k) = -f(k - 2t²). So, the inequality becomes:f(t² - 2t) - f(k - 2t²) < 0 => f(t² - 2t) < f(k - 2t²).Since f is strictly decreasing, this implies t² - 2t > k - 2t².Yes, that's correct.Then, 3t² - 2t - k > 0 for all t.Quadratic in t: 3t² - 2t - k.Since 3 > 0, it opens upwards. For it to be always positive, discriminant must be negative.Discriminant D = (-2)^2 - 4*3*(-k) = 4 + 12k.Set D < 0: 4 + 12k < 0 => 12k < -4 => k < -1/3.Yes, that seems correct.Therefore, the range of k is k < -1/3.I think that's it.**Final Answer**(I) The values are boxed{a = 1} and boxed{b = 1}.(II) The range of k is boxed{k < -dfrac{1}{3}}.

answer：Okay, so I have this geometry problem here about three circles. Let me try to visualize it first. There's a big circle ( k ) with radius 9. Inside it, there are two smaller circles, ( k_1 ) and ( k_2 ), with radii 6 and 3 respectively. These two smaller circles are externally tangent to each other, which means they touch at exactly one point but don't overlap. Also, both of these smaller circles touch the big circle ( k ) from the inside. The problem is asking for the length of the common external tangent ( PQ ) of ( k_1 ) and ( k_2 ) that intersects the big circle ( k ) at points ( P ) and ( Q ). Hmm, okay. So I need to find the length of this chord ( PQ ) in the big circle.First, I think it would help to draw a diagram. Let me sketch the big circle ( k ) with center ( O ) and radius 9. Inside it, circle ( k_1 ) with radius 6 is tangent to ( k ) at some point, and circle ( k_2 ) with radius 3 is also tangent to ( k ) at another point. Since ( k_1 ) and ( k_2 ) are externally tangent, the distance between their centers should be equal to the sum of their radii, which is ( 6 + 3 = 9 ).Wait, but both ( k_1 ) and ( k_2 ) are inside ( k ), so their centers must be located somewhere inside ( k ). Let me denote the centers of ( k_1 ), ( k_2 ), and ( k ) as ( O_1 ), ( O_2 ), and ( O ) respectively. Since ( k_1 ) and ( k_2 ) are tangent to ( k ), the distance from ( O ) to ( O_1 ) should be ( 9 - 6 = 3 ), and the distance from ( O ) to ( O_2 ) should be ( 9 - 3 = 6 ).Wait, that doesn't make sense. If ( O_1 ) is 3 units away from ( O ) and ( O_2 ) is 6 units away, and the distance between ( O_1 ) and ( O_2 ) is 9 units, then ( O ), ( O_1 ), and ( O_2 ) must be colinear. Because ( 3 + 6 = 9 ), which is the distance between ( O_1 ) and ( O_2 ). So, the centers lie on a straight line, with ( O ) somewhere in between.Let me confirm that. If ( O_1 ) is 3 units from ( O ) and ( O_2 ) is 6 units from ( O ), then the distance between ( O_1 ) and ( O_2 ) is ( 3 + 6 = 9 ), which matches the sum of their radii. So yes, they are colinear.So, let's set up a coordinate system to make things easier. Let me place the center ( O ) of the big circle ( k ) at the origin (0,0). Then, since ( O_1 ) is 3 units away from ( O ) and ( O_2 ) is 6 units away, and they are colinear, I can place ( O_1 ) at (3,0) and ( O_2 ) at (-6,0). Wait, no, that would make the distance between ( O_1 ) and ( O_2 ) equal to 9 units, which is correct. But hold on, if ( O_1 ) is at (3,0) and ( O_2 ) is at (-6,0), then the distance between them is indeed 9 units. But in this case, ( O_1 ) is inside ( k ), which has radius 9, so (3,0) is 3 units from the center, which is correct because ( k_1 ) has radius 6, so the distance from ( O ) to ( O_1 ) is ( 9 - 6 = 3 ). Similarly, ( O_2 ) is 6 units from ( O ), which is ( 9 - 3 = 6 ). So that seems correct.Wait, but in this setup, ( O_1 ) is at (3,0) and ( O_2 ) is at (-6,0). So, the common external tangent of ( k_1 ) and ( k_2 ) would be a line that touches both circles without crossing between them. Since they are on opposite sides of the center ( O ), the external tangent would be somewhere above or below the line connecting their centers.But in this case, since ( O_1 ) is at (3,0) and ( O_2 ) is at (-6,0), the line connecting their centers is along the x-axis. So, the external tangent would be a line above or below the x-axis that touches both circles.Wait, but in the problem statement, it says that the common external tangent intersects ( k ) at ( P ) and ( Q ). So, the tangent line is not just a tangent to ( k_1 ) and ( k_2 ), but it also intersects the big circle ( k ). So, the tangent line is a chord of ( k ), and we need to find the length of that chord.Okay, so I need to find the equation of the common external tangent to ( k_1 ) and ( k_2 ), then find where this line intersects the big circle ( k ), and then compute the distance between those two intersection points ( P ) and ( Q ).Let me recall how to find the equation of a common external tangent to two circles. The general approach is to consider the two circles and find a line that touches both without crossing the line segment joining their centers.Given two circles with centers ( O_1 ) and ( O_2 ) and radii ( r_1 ) and ( r_2 ), the distance between the centers is ( d ). For external tangents, the angle ( theta ) between the line connecting the centers and the tangent line can be found using similar triangles.In this case, ( O_1 ) is at (3,0) with radius 6, and ( O_2 ) is at (-6,0) with radius 3. The distance between ( O_1 ) and ( O_2 ) is 9 units. So, ( d = 9 ), ( r_1 = 6 ), ( r_2 = 3 ).The formula for the distance between the centers along the line connecting them is ( d ). The length of the external tangent between two circles is given by ( sqrt{d^2 - (r_1 + r_2)^2} ). Wait, no, that's for the internal tangent. For external tangent, it's ( sqrt{d^2 - (r_1 - r_2)^2} ). Wait, actually, let me think carefully.No, wait, actually, for external tangent, the formula is ( sqrt{d^2 - (r_1 - r_2)^2} ), and for internal tangent, it's ( sqrt{d^2 - (r_1 + r_2)^2} ). But in this case, since the circles are externally tangent, the external tangent lines are the ones that don't cross between the circles, so the formula should be applicable.But in our case, the circles are externally tangent, so the distance between centers is equal to the sum of radii, which is 9. Wait, ( r_1 + r_2 = 6 + 3 = 9 ), which is equal to the distance between centers. So, that means the circles are externally tangent, so they have exactly one common tangent at the point where they touch. But in the problem statement, it says "a common external tangent", which suggests that there is more than one, but since they are externally tangent, they have three common tangents: two external and one at the point of contact.Wait, no, actually, when two circles are externally tangent, they have three common tangent lines: two external ones that don't pass through the point of contact, and one common tangent at the point where they touch. So, in this case, the external tangent that is not at the point of contact would be the one we're interested in.But wait, in our case, the circles are externally tangent, so the distance between centers is equal to the sum of radii, which is 9. So, if I try to compute the length of the external tangent, it would be ( sqrt{d^2 - (r_1 - r_2)^2} ). Plugging in the values, ( d = 9 ), ( r_1 = 6 ), ( r_2 = 3 ), so ( r_1 - r_2 = 3 ). Therefore, the length of the external tangent is ( sqrt{9^2 - 3^2} = sqrt{81 - 9} = sqrt{72} = 6sqrt{2} ).Wait, but that's the length of the external tangent between the two circles. But in our problem, the tangent line intersects the big circle ( k ) at points ( P ) and ( Q ). So, we need to find the length of chord ( PQ ) in circle ( k ).So, perhaps we can find the equation of the external tangent, then find its intersection points with circle ( k ), and then compute the distance between those points.Let me try to find the equation of the external tangent. Since the centers ( O_1 ) and ( O_2 ) are on the x-axis at (3,0) and (-6,0), respectively, the external tangent will be symmetric with respect to the x-axis. So, there will be two external tangents, one above the x-axis and one below. Since the problem mentions a common external tangent, I think we can choose either one, as they will be symmetric, and the length ( PQ ) will be the same.Let me focus on the upper external tangent. To find its equation, I can use the fact that the tangent line will touch ( k_1 ) at some point ( T_1 ) and ( k_2 ) at some point ( T_2 ). The line ( O_1T_1 ) is perpendicular to the tangent line, and similarly, ( O_2T_2 ) is perpendicular to the tangent line.So, if I can find the slope of the tangent line, I can write its equation. Let me denote the slope as ( m ). Then, the slope of the line connecting ( O_1 ) to ( T_1 ) is ( -1/m ), since it's perpendicular.Let me consider the coordinates. Let me denote the tangent line as ( y = mx + c ). Since it's an external tangent above the x-axis, ( c ) will be positive.The distance from ( O_1 ) (3,0) to the tangent line must be equal to the radius of ( k_1 ), which is 6. Similarly, the distance from ( O_2 ) (-6,0) to the tangent line must be equal to the radius of ( k_2 ), which is 3.The formula for the distance from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ) is ( frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}} ). In our case, the tangent line is ( y = mx + c ), which can be rewritten as ( mx - y + c = 0 ). So, ( a = m ), ( b = -1 ), ( c = c ).So, the distance from ( O_1 ) (3,0) to the tangent line is:( frac{|m*3 - 1*0 + c|}{sqrt{m^2 + 1}}} = frac{|3m + c|}{sqrt{m^2 + 1}}} = 6 )Similarly, the distance from ( O_2 ) (-6,0) to the tangent line is:( frac{|m*(-6) - 1*0 + c|}{sqrt{m^2 + 1}}} = frac{|-6m + c|}{sqrt{m^2 + 1}}} = 3 )So, we have two equations:1. ( |3m + c| = 6sqrt{m^2 + 1} )2. ( |-6m + c| = 3sqrt{m^2 + 1} )Since the tangent is above the x-axis, and given the positions of ( O_1 ) and ( O_2 ), I think ( c ) will be positive, and the expressions inside the absolute value will be positive as well. So, we can drop the absolute value:1. ( 3m + c = 6sqrt{m^2 + 1} )2. ( -6m + c = 3sqrt{m^2 + 1} )Now, we have a system of two equations:Equation (1): ( 3m + c = 6sqrt{m^2 + 1} )Equation (2): ( -6m + c = 3sqrt{m^2 + 1} )Let me subtract Equation (2) from Equation (1):( (3m + c) - (-6m + c) = 6sqrt{m^2 + 1} - 3sqrt{m^2 + 1} )Simplify:( 3m + c + 6m - c = 3sqrt{m^2 + 1} )( 9m = 3sqrt{m^2 + 1} )Divide both sides by 3:( 3m = sqrt{m^2 + 1} )Now, square both sides:( 9m^2 = m^2 + 1 )Subtract ( m^2 + 1 ) from both sides:( 8m^2 - 1 = 0 )So,( 8m^2 = 1 )( m^2 = frac{1}{8} )( m = pm frac{1}{2sqrt{2}} = pm frac{sqrt{2}}{4} )Since we are considering the upper external tangent, which has a positive slope, we take ( m = frac{sqrt{2}}{4} ).Now, plug this back into Equation (2) to find ( c ):( -6m + c = 3sqrt{m^2 + 1} )First, compute ( sqrt{m^2 + 1} ):( m^2 = frac{1}{8} ), so ( m^2 + 1 = frac{1}{8} + 1 = frac{9}{8} ), so ( sqrt{frac{9}{8}} = frac{3}{2sqrt{2}} = frac{3sqrt{2}}{4} )So,( -6*frac{sqrt{2}}{4} + c = 3*frac{3sqrt{2}}{4} )Simplify:( -frac{6sqrt{2}}{4} + c = frac{9sqrt{2}}{4} )Multiply numerator and denominator:( -frac{3sqrt{2}}{2} + c = frac{9sqrt{2}}{4} )Add ( frac{3sqrt{2}}{2} ) to both sides:( c = frac{9sqrt{2}}{4} + frac{3sqrt{2}}{2} )Convert ( frac{3sqrt{2}}{2} ) to quarters:( frac{3sqrt{2}}{2} = frac{6sqrt{2}}{4} )So,( c = frac{9sqrt{2}}{4} + frac{6sqrt{2}}{4} = frac{15sqrt{2}}{4} )Therefore, the equation of the tangent line is:( y = frac{sqrt{2}}{4}x + frac{15sqrt{2}}{4} )Alternatively, we can write it as:( y = frac{sqrt{2}}{4}(x + 15) )Wait, let me check that. If I factor out ( frac{sqrt{2}}{4} ), it's ( frac{sqrt{2}}{4}x + frac{15sqrt{2}}{4} = frac{sqrt{2}}{4}(x + 15) ). Yes, that's correct.Now, we need to find where this line intersects the big circle ( k ) with center at (0,0) and radius 9. The equation of circle ( k ) is:( x^2 + y^2 = 81 )Substitute ( y = frac{sqrt{2}}{4}(x + 15) ) into the circle equation:( x^2 + left( frac{sqrt{2}}{4}(x + 15) right)^2 = 81 )Let me compute ( left( frac{sqrt{2}}{4}(x + 15) right)^2 ):( left( frac{sqrt{2}}{4} right)^2 (x + 15)^2 = frac{2}{16}(x + 15)^2 = frac{1}{8}(x + 15)^2 )So, the equation becomes:( x^2 + frac{1}{8}(x + 15)^2 = 81 )Multiply both sides by 8 to eliminate the denominator:( 8x^2 + (x + 15)^2 = 648 )Expand ( (x + 15)^2 ):( x^2 + 30x + 225 )So, the equation becomes:( 8x^2 + x^2 + 30x + 225 = 648 )Combine like terms:( 9x^2 + 30x + 225 = 648 )Subtract 648 from both sides:( 9x^2 + 30x + 225 - 648 = 0 )Simplify:( 9x^2 + 30x - 423 = 0 )Divide the entire equation by 3 to simplify:( 3x^2 + 10x - 141 = 0 )Now, we can solve this quadratic equation for ( x ) using the quadratic formula:( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Where ( a = 3 ), ( b = 10 ), and ( c = -141 ):Compute discriminant ( D ):( D = b^2 - 4ac = 10^2 - 4*3*(-141) = 100 + 1692 = 1792 )So,( x = frac{-10 pm sqrt{1792}}{6} )Simplify ( sqrt{1792} ):1792 = 256 * 7, because 256 * 7 = 1792. So,( sqrt{1792} = sqrt{256 * 7} = 16sqrt{7} )Therefore,( x = frac{-10 pm 16sqrt{7}}{6} )Simplify fractions:( x = frac{-5 pm 8sqrt{7}}{3} )So, the x-coordinates of points ( P ) and ( Q ) are ( frac{-5 + 8sqrt{7}}{3} ) and ( frac{-5 - 8sqrt{7}}{3} ).Now, let's find the corresponding y-coordinates using the equation of the tangent line:( y = frac{sqrt{2}}{4}(x + 15) )For ( x = frac{-5 + 8sqrt{7}}{3} ):( y = frac{sqrt{2}}{4}left( frac{-5 + 8sqrt{7}}{3} + 15 right) )Convert 15 to thirds:( 15 = frac{45}{3} )So,( y = frac{sqrt{2}}{4}left( frac{-5 + 8sqrt{7} + 45}{3} right) = frac{sqrt{2}}{4}left( frac{40 + 8sqrt{7}}{3} right) = frac{sqrt{2}}{4} * frac{8(5 + sqrt{7})}{3} = frac{2sqrt{2}(5 + sqrt{7})}{3} )Similarly, for ( x = frac{-5 - 8sqrt{7}}{3} ):( y = frac{sqrt{2}}{4}left( frac{-5 - 8sqrt{7}}{3} + 15 right) = frac{sqrt{2}}{4}left( frac{-5 - 8sqrt{7} + 45}{3} right) = frac{sqrt{2}}{4}left( frac{40 - 8sqrt{7}}{3} right) = frac{sqrt{2}}{4} * frac{8(5 - sqrt{7})}{3} = frac{2sqrt{2}(5 - sqrt{7})}{3} )So, the coordinates of points ( P ) and ( Q ) are:( P left( frac{-5 + 8sqrt{7}}{3}, frac{2sqrt{2}(5 + sqrt{7})}{3} right) )and( Q left( frac{-5 - 8sqrt{7}}{3}, frac{2sqrt{2}(5 - sqrt{7})}{3} right) )Now, to find the distance ( PQ ), we can use the distance formula:( PQ = sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2} )Compute ( x_P - x_Q ):( frac{-5 + 8sqrt{7}}{3} - frac{-5 - 8sqrt{7}}{3} = frac{(-5 + 8sqrt{7}) - (-5 - 8sqrt{7})}{3} = frac{16sqrt{7}}{3} )Compute ( y_P - y_Q ):( frac{2sqrt{2}(5 + sqrt{7})}{3} - frac{2sqrt{2}(5 - sqrt{7})}{3} = frac{2sqrt{2}(5 + sqrt{7} - 5 + sqrt{7})}{3} = frac{2sqrt{2}(2sqrt{7})}{3} = frac{4sqrt{14}}{3} )Now, plug these into the distance formula:( PQ = sqrt{left( frac{16sqrt{7}}{3} right)^2 + left( frac{4sqrt{14}}{3} right)^2} )Compute each term:( left( frac{16sqrt{7}}{3} right)^2 = frac{256 * 7}{9} = frac{1792}{9} )( left( frac{4sqrt{14}}{3} right)^2 = frac{16 * 14}{9} = frac{224}{9} )Add them together:( frac{1792}{9} + frac{224}{9} = frac{2016}{9} = 224 )So,( PQ = sqrt{224} = sqrt{16 * 14} = 4sqrt{14} )Therefore, the length of ( PQ ) is ( 4sqrt{14} ).**Final Answer**The length of ( PQ ) is (boxed{4sqrt{14}}).

answer：First, I need to identify the like terms in the expression. The terms (3y), (4y), and (5y) all contain the variable (y), so they are like terms and can be combined.Next, I'll add the coefficients of these like terms:[3 + 4 + 5 = 12]This gives me the combined term (12y).Finally, I'll add the constant term (7) to the combined variable term:[12y + 7]Therefore, the simplified form of the expression is (boxed{12y + 7}).

answer：Okay, so I have this problem here: the conic curve is given by the equation ( x^2 + m y^2 = 1 ). The focus of this conic lies on the x-axis, and the eccentricity is ( frac{1}{2} ). I need to find the value of the real number ( m ).Hmm, let me start by recalling what I know about conic sections. The general form of a conic section is ( Ax^2 + By^2 + Cx + Dy + E = 0 ). In this case, the equation is ( x^2 + m y^2 = 1 ), which looks like an ellipse or a hyperbola depending on the coefficients.Since the focus lies on the x-axis, that suggests that the major axis of the conic is along the x-axis. If it were a hyperbola, the foci would still be on the x-axis, but the eccentricity of a hyperbola is greater than 1, and here the eccentricity is ( frac{1}{2} ), which is less than 1. So, it must be an ellipse.Alright, so it's an ellipse. The standard form of an ellipse centered at the origin with major axis along the x-axis is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a > b ). Comparing this with the given equation ( x^2 + m y^2 = 1 ), I can rewrite the given equation as ( frac{x^2}{1} + frac{y^2}{1/m} = 1 ). So, that means ( a^2 = 1 ) and ( b^2 = frac{1}{m} ).Now, the eccentricity ( e ) of an ellipse is given by ( e = sqrt{1 - frac{b^2}{a^2}} ). Plugging in the known values, we have ( e = sqrt{1 - frac{1/m}{1}} = sqrt{1 - frac{1}{m}} ).We are told that the eccentricity is ( frac{1}{2} ). So, setting up the equation:[frac{1}{2} = sqrt{1 - frac{1}{m}}]To solve for ( m ), I'll square both sides to eliminate the square root:[left( frac{1}{2} right)^2 = 1 - frac{1}{m}][frac{1}{4} = 1 - frac{1}{m}]Now, solving for ( frac{1}{m} ):[frac{1}{m} = 1 - frac{1}{4}][frac{1}{m} = frac{3}{4}]Taking the reciprocal of both sides to solve for ( m ):[m = frac{4}{3}]So, the value of ( m ) is ( frac{4}{3} ).Wait a second, let me double-check my steps to make sure I didn't make a mistake. Starting from the equation ( x^2 + m y^2 = 1 ), I correctly identified it as an ellipse because the eccentricity is less than 1. Then, I compared it to the standard form and found ( a^2 = 1 ) and ( b^2 = frac{1}{m} ). The formula for eccentricity is correct, and plugging in the values led me to the equation ( frac{1}{2} = sqrt{1 - frac{1}{m}} ). Squaring both sides gives ( frac{1}{4} = 1 - frac{1}{m} ), which simplifies to ( frac{1}{m} = frac{3}{4} ), so ( m = frac{4}{3} ). That seems right.I think I'm confident with this answer. It makes sense because if ( m ) were 1, the equation would be a circle, but since ( m ) is greater than 1, it's an ellipse elongated along the x-axis, which aligns with the focus being on the x-axis. The eccentricity being ( frac{1}{2} ) also fits because it's a measure of how elongated the ellipse is, and ( frac{1}{2} ) is a moderate value, not too close to 0 or 1.So, yeah, I think ( m = frac{4}{3} ) is the correct answer.

answer：Alright, so I've got this geometry problem here about a cube, and I need to find the volume of a specific pyramid. Let me try to visualize the cube and all the points mentioned.First, the cube is labeled ABCDA'B'C'D'. I remember that in such cubes, the vertices with primes are the ones directly above the non-prime vertices. So, A is connected to A', B to B', and so on. The edge length is 2, which is good to know because it will help with calculations later.The point K is the center of edge AB. Since AB is an edge of length 2, K must be exactly halfway between A and B. So, if I imagine the cube, K is right in the middle of the bottom front edge.Now, the problem says that the plane containing points B', D', and K intersects the edge AD at point L. I need to find where exactly L is on AD. Once I have L, I can figure out the shape of the base of the pyramid, which is the quadrilateral D'B'KL.To find L, I think I need to determine the equation of the plane passing through B', D', and K. Then, I can find where this plane intersects the edge AD. Let me recall how to find the equation of a plane given three points.First, I should assign coordinates to all the vertices to make this easier. Let's place the cube in a 3D coordinate system where point A is at (0, 0, 0). Then, since the edge length is 2, the coordinates of the other points can be determined as follows:- A = (0, 0, 0)- B = (2, 0, 0)- C = (2, 2, 0)- D = (0, 2, 0)- A' = (0, 0, 2)- B' = (2, 0, 2)- C' = (2, 2, 2)- D' = (0, 2, 2)Point K is the midpoint of AB, so its coordinates are the average of A and B:K = ((0 + 2)/2, (0 + 0)/2, (0 + 0)/2) = (1, 0, 0)Now, I have the coordinates of B', D', and K:- B' = (2, 0, 2)- D' = (0, 2, 2)- K = (1, 0, 0)I need to find the equation of the plane passing through these three points. To do this, I can use the general equation of a plane: ax + by + cz + d = 0. Since the plane passes through B', D', and K, substituting these points into the equation will give me a system of equations to solve for a, b, c, and d.Let's plug in B' = (2, 0, 2):2a + 0*b + 2c + d = 0 => 2a + 2c + d = 0 ...(1)Plug in D' = (0, 2, 2):0*a + 2b + 2c + d = 0 => 2b + 2c + d = 0 ...(2)Plug in K = (1, 0, 0):1a + 0*b + 0*c + d = 0 => a + d = 0 ...(3)Now, I have three equations:1) 2a + 2c + d = 02) 2b + 2c + d = 03) a + d = 0Let me solve equation (3) first: a = -dSubstitute a = -d into equation (1):2*(-d) + 2c + d = 0 => -2d + 2c + d = 0 => -d + 2c = 0 => 2c = d => c = d/2Similarly, substitute a = -d into equation (2):2b + 2c + d = 0But we know c = d/2, so:2b + 2*(d/2) + d = 0 => 2b + d + d = 0 => 2b + 2d = 0 => b = -dSo now, we have:a = -db = -dc = d/2We can choose d to be any value except zero (since the plane equation can't be all zeros). Let's choose d = 2 for simplicity.Then:a = -2b = -2c = 1So, the equation of the plane is:-2x - 2y + z + 2 = 0I can write this as:2x + 2y - z = 2Okay, so that's the equation of the plane containing B', D', and K.Now, I need to find where this plane intersects edge AD. Edge AD goes from A = (0, 0, 0) to D = (0, 2, 0). So, any point on AD can be parameterized as (0, t, 0) where t ranges from 0 to 2.Let me substitute x = 0, z = 0 into the plane equation:2*0 + 2y - 0 = 2 => 2y = 2 => y = 1So, the point L is at (0, 1, 0). That makes sense; it's halfway up edge AD.Now, the base of the pyramid is the quadrilateral D'B'KL. Let me list the coordinates of these points:- D' = (0, 2, 2)- B' = (2, 0, 2)- K = (1, 0, 0)- L = (0, 1, 0)So, the quadrilateral is a four-sided figure connecting these points. To find the volume of the pyramid with apex A and base D'B'KL, I need to calculate the area of the base and then find the height from A to the base.But wait, actually, since it's a pyramid, the volume can be found using the scalar triple product of vectors from the apex to the base points. Alternatively, I can divide the base into triangles and calculate the volume accordingly.Alternatively, another method is to use coordinates and the formula for the volume of a pyramid with a polygonal base.But perhaps the most straightforward way is to use the determinant method. Since I have coordinates for all the points, I can set up vectors from A to each of the base points and compute the volume using the scalar triple product.Let me recall that the volume of a pyramid is (1/6) times the absolute value of the scalar triple product of three edges meeting at the apex.But in this case, the base is a quadrilateral, so maybe I need to split it into two triangles and compute the volume for each, then add them up.Yes, that sounds manageable.So, let's split the quadrilateral D'B'KL into two triangles: D'B'K and D'KL.First, let's compute the volume contributed by triangle D'B'K.The vectors from A to these points are:- AD' = D' - A = (0, 2, 2)- AB' = B' - A = (2, 0, 2)- AK = K - A = (1, 0, 0)Wait, but actually, for the scalar triple product, I need vectors from the apex A to three points defining the base. Since the base is a triangle, I can use vectors AD', AB', and AK.But wait, triangle D'B'K is formed by points D', B', and K. So, vectors from A to these points are AD', AB', and AK.But actually, the scalar triple product is for three vectors emanating from the same point. So, if I take vectors AD', AB', and AK, their scalar triple product will give the volume of the parallelepiped, and (1/6) of that is the volume of the tetrahedron.But in this case, the base is triangle D'B'K, so the volume of the pyramid with base D'B'K and apex A is (1/3)*Area of base * height.Wait, maybe I should compute the area of triangle D'B'K and then find the height from A to this triangle.But that might be more complicated. Alternatively, using the scalar triple product for the tetrahedron.Wait, perhaps it's better to use coordinates and set up the determinant.Let me recall that the volume of a tetrahedron with vertices at points P, Q, R, S is (1/6)*|det(B - A, C - A, D - A)|, where A, B, C, D are the points.But in this case, the base is a quadrilateral, so maybe I need to split it into two tetrahedrons.Yes, that's a good approach. So, the pyramid with base D'B'KL can be split into two tetrahedrons: one with base D'B'K and apex A, and another with base D'KL and apex A.So, I can compute the volume of each tetrahedron and add them together.Let me compute the volume of tetrahedron AD'B'K first.The coordinates are:A = (0, 0, 0)D' = (0, 2, 2)B' = (2, 0, 2)K = (1, 0, 0)So, vectors from A are:AD' = (0, 2, 2)AB' = (2, 0, 2)AK = (1, 0, 0)The scalar triple product is AD' ⋅ (AB' × AK)First, compute AB' × AK:AB' = (2, 0, 2)AK = (1, 0, 0)Cross product:|i   j   k||2   0   2||1   0   0|= i*(0*0 - 2*0) - j*(2*0 - 2*1) + k*(2*0 - 0*1)= i*(0) - j*(-2) + k*(0)= (0, 2, 0)Now, AD' ⋅ (0, 2, 0) = (0, 2, 2) ⋅ (0, 2, 0) = 0*0 + 2*2 + 2*0 = 4So, the volume of tetrahedron AD'B'K is (1/6)*|4| = 4/6 = 2/3Now, let's compute the volume of tetrahedron AD'KL.Points:A = (0, 0, 0)D' = (0, 2, 2)K = (1, 0, 0)L = (0, 1, 0)Vectors from A:AD' = (0, 2, 2)AK = (1, 0, 0)AL = (0, 1, 0)The scalar triple product is AD' ⋅ (AK × AL)First, compute AK × AL:AK = (1, 0, 0)AL = (0, 1, 0)Cross product:|i   j   k||1   0   0||0   1   0|= i*(0*0 - 0*1) - j*(1*0 - 0*0) + k*(1*1 - 0*0)= (0, 0, 1)Now, AD' ⋅ (0, 0, 1) = (0, 2, 2) ⋅ (0, 0, 1) = 0*0 + 2*0 + 2*1 = 2So, the volume of tetrahedron AD'KL is (1/6)*|2| = 2/6 = 1/3Adding the two volumes together: 2/3 + 1/3 = 1So, the total volume of the pyramid is 1.Wait, that seems straightforward. Let me just double-check my calculations to make sure I didn't make any mistakes.For the first tetrahedron AD'B'K:- Vectors: AD' = (0,2,2), AB' = (2,0,2), AK = (1,0,0)- Cross product AB' × AK = (0,2,0)- Dot product with AD' = 4- Volume = 4/6 = 2/3For the second tetrahedron AD'KL:- Vectors: AD' = (0,2,2), AK = (1,0,0), AL = (0,1,0)- Cross product AK × AL = (0,0,1)- Dot product with AD' = 2- Volume = 2/6 = 1/3Total volume: 2/3 + 1/3 = 1Yes, that seems correct. So, the volume of the pyramid is 1.

answer：First, I need to identify the cost per meter for each side of the pentagonal field. The first two sides are priced at Rs. 7 per meter, the middle side at Rs. 5 per meter, and the last two sides at Rs. 10 per meter.Next, I'll calculate the cost for each side individually by multiplying the length of each side by its respective cost per meter.Finally, I'll sum up the costs of all five sides to determine the total fencing cost for the entire pentagonal field.