answer：Okay, so I have this problem where I need to find real numbers x, y, z, all greater than 4, that satisfy the equation:[frac{(x + 3)^2}{y + z - 3} + frac{(y + 5)^2}{z + x - 5} + frac{(z + 7)^2}{x + y - 7} = 45.]Hmm, this looks a bit complicated, but maybe I can use some inequality to simplify it. I remember that the Cauchy-Schwarz inequality is often useful in these kinds of problems. Let me recall the Cauchy-Schwarz inequality in the form:[left( sum frac{a_i^2}{b_i} right) left( sum b_i right) geq left( sum a_i right)^2.]Yes, that seems right. So, in this problem, each term in the sum is of the form (frac{(something)^2}{something}). Maybe I can apply Cauchy-Schwarz here.Let me set:- (a_1 = x + 3), (a_2 = y + 5), (a_3 = z + 7)- (b_1 = y + z - 3), (b_2 = z + x - 5), (b_3 = x + y - 7)So, the given equation is:[frac{a_1^2}{b_1} + frac{a_2^2}{b_2} + frac{a_3^2}{b_3} = 45.]Applying Cauchy-Schwarz, I get:[left( frac{a_1^2}{b_1} + frac{a_2^2}{b_2} + frac{a_3^2}{b_3} right) left( b_1 + b_2 + b_3 right) geq (a_1 + a_2 + a_3)^2.]Plugging in the values, this becomes:[45 times (b_1 + b_2 + b_3) geq (a_1 + a_2 + a_3)^2.]Now, let me compute (b_1 + b_2 + b_3):- (b_1 = y + z - 3)- (b_2 = z + x - 5)- (b_3 = x + y - 7)Adding them up:[b_1 + b_2 + b_3 = (y + z - 3) + (z + x - 5) + (x + y - 7) = 2x + 2y + 2z - 15.]So, (b_1 + b_2 + b_3 = 2(x + y + z) - 15). Let me denote (s = x + y + z). Then, (b_1 + b_2 + b_3 = 2s - 15).Similarly, (a_1 + a_2 + a_3 = (x + 3) + (y + 5) + (z + 7) = x + y + z + 15 = s + 15).So, substituting back into the inequality:[45(2s - 15) geq (s + 15)^2.]Let me expand both sides:Left side: (45 times (2s - 15) = 90s - 675).Right side: ((s + 15)^2 = s^2 + 30s + 225).So, the inequality becomes:[90s - 675 geq s^2 + 30s + 225.]Let me bring all terms to one side:[0 geq s^2 + 30s + 225 - 90s + 675.]Simplify the terms:- (30s - 90s = -60s)- (225 + 675 = 900)So, the inequality is:[0 geq s^2 - 60s + 900.]Which can be rewritten as:[s^2 - 60s + 900 leq 0.]Hmm, this is a quadratic in s. Let me see if I can factor it or find its roots.The quadratic is (s^2 - 60s + 900). Let me compute the discriminant:Discriminant (D = (-60)^2 - 4 times 1 times 900 = 3600 - 3600 = 0).So, the quadratic has a double root. The root is:[s = frac{60}{2} = 30.]Therefore, the quadratic can be written as ((s - 30)^2 leq 0).Since a square is always non-negative, the only solution is (s = 30).So, (x + y + z = 30).Now, going back to the Cauchy-Schwarz inequality, since equality holds, the condition for equality in Cauchy-Schwarz must be satisfied. That is, the ratios (frac{a_1}{b_1} = frac{a_2}{b_2} = frac{a_3}{b_3}) must be equal.Let me denote this common ratio as k. So,[frac{x + 3}{y + z - 3} = frac{y + 5}{z + x - 5} = frac{z + 7}{x + y - 7} = k.]Since (x + y + z = 30), I can express each denominator in terms of s.For example:- (y + z - 3 = (30 - x) - 3 = 27 - x)- (z + x - 5 = (30 - y) - 5 = 25 - y)- (x + y - 7 = (30 - z) - 7 = 23 - z)So, the equations become:1. (frac{x + 3}{27 - x} = k)2. (frac{y + 5}{25 - y} = k)3. (frac{z + 7}{23 - z} = k)So, each of these is equal to k. Let me solve each equation for x, y, z.From equation 1:[x + 3 = k(27 - x)][x + 3 = 27k - kx][x + kx = 27k - 3][x(1 + k) = 27k - 3][x = frac{27k - 3}{1 + k}]Similarly, from equation 2:[y + 5 = k(25 - y)][y + 5 = 25k - ky][y + ky = 25k - 5][y(1 + k) = 25k - 5][y = frac{25k - 5}{1 + k}]From equation 3:[z + 7 = k(23 - z)][z + 7 = 23k - kz][z + kz = 23k - 7][z(1 + k) = 23k - 7][z = frac{23k - 7}{1 + k}]So, now I have expressions for x, y, z in terms of k. Since x + y + z = 30, let me add these expressions:[x + y + z = frac{27k - 3}{1 + k} + frac{25k - 5}{1 + k} + frac{23k - 7}{1 + k} = 30]Combine the numerators:[frac{(27k - 3) + (25k - 5) + (23k - 7)}{1 + k} = 30]Compute the numerator:- 27k + 25k + 23k = 75k- -3 -5 -7 = -15So,[frac{75k - 15}{1 + k} = 30]Multiply both sides by (1 + k):[75k - 15 = 30(1 + k)][75k - 15 = 30 + 30k][75k - 30k = 30 + 15][45k = 45][k = 1]So, k is 1. Now, substitute k = 1 back into the expressions for x, y, z.For x:[x = frac{27(1) - 3}{1 + 1} = frac{27 - 3}{2} = frac{24}{2} = 12]For y:[y = frac{25(1) - 5}{1 + 1} = frac{25 - 5}{2} = frac{20}{2} = 10]For z:[z = frac{23(1) - 7}{1 + 1} = frac{23 - 7}{2} = frac{16}{2} = 8]So, x = 12, y = 10, z = 8.Let me check if these satisfy the original equation.First, compute each term:1. (frac{(x + 3)^2}{y + z - 3}) - x + 3 = 15 - y + z - 3 = 10 + 8 - 3 = 15 - So, term = (15^2 / 15 = 225 / 15 = 15)2. (frac{(y + 5)^2}{z + x - 5}) - y + 5 = 15 - z + x - 5 = 8 + 12 - 5 = 15 - Term = (15^2 / 15 = 225 / 15 = 15)3. (frac{(z + 7)^2}{x + y - 7}) - z + 7 = 15 - x + y - 7 = 12 + 10 - 7 = 15 - Term = (15^2 / 15 = 225 / 15 = 15)Adding them up: 15 + 15 + 15 = 45, which matches the given equation.Also, all x, y, z are greater than 4, as required.So, the ordered triple is (12, 10, 8).**Final Answer**The ordered triple is boxed{(12, 10, 8)}.

answer：First, I need to determine the set ( A ) by solving the inequality ( x^2 + 2x - 3 < 0 ). To do this, I'll factor the quadratic expression.The quadratic ( x^2 + 2x - 3 ) can be factored as ( (x + 3)(x - 1) ). Setting this less than zero gives:[(x + 3)(x - 1) < 0]To find the solution, I'll identify the critical points where the expression equals zero, which are ( x = -3 ) and ( x = 1 ). These points divide the number line into intervals. Testing each interval:- For ( x < -3 ), the product is positive.- For ( -3 < x < 1 ), the product is negative.- For ( x > 1 ), the product is positive.Since we're looking for where the expression is less than zero, the solution is:[A = (-3, 1)]Next, I'll determine the set ( B ) from the inequality ( x - a < 0 ), which simplifies to:[x < a]So, ( B = (-infty, a) ).Given that ( A subseteq B ), every element of ( A ) must also be in ( B ). This means the upper bound of ( A ), which is ( 1 ), must be less than or equal to ( a ):[1 leq a]Therefore, the range of values for ( a ) is ( a geq 1 ).

answer：First, I need to solve the equation (frac{4x - 2}{5x - 5} = frac{3}{4}). To eliminate the fractions, I'll multiply both sides of the equation by (5x - 5) and by 4.This gives me:[4(4x - 2) = 3(5x - 5)]Next, I'll expand both sides of the equation:[16x - 8 = 15x - 15]To isolate (x), I'll subtract (15x) from both sides:[16x - 15x - 8 = -15][x - 8 = -15]Finally, I'll add 8 to both sides to solve for (x):[x = -15 + 8][x = -7]

answer：First, I need to find the midpoint of segment ( s_1 ) with endpoints at ( (3, -2) ) and ( (-7, 4) ). To do this, I'll use the midpoint formula, which averages the x-coordinates and the y-coordinates of the endpoints.Calculating the x-coordinate of the midpoint:[frac{3 + (-7)}{2} = frac{-4}{2} = -2]Calculating the y-coordinate of the midpoint:[frac{-2 + 4}{2} = frac{2}{2} = 1]So, the midpoint of ( s_1 ) is ( (-2, 1) ).Next, I need to translate this midpoint by 3 units to the right and 2 units down to find the midpoint of ( s_2 ). Translating 3 units to the right increases the x-coordinate by 3, and translating 2 units down decreases the y-coordinate by 2.Translating the x-coordinate:[-2 + 3 = 1]Translating the y-coordinate:[1 - 2 = -1]Therefore, the midpoint of segment ( s_2 ) is ( (1, -1) ).

answer：Alright, so I have this problem where an ancient British merchant had a 15-pound weight that shattered into 4 pieces. Each piece weighs an integer number of pounds, and these pieces can be used to measure any integer weight from 1 to 15 pounds using a balance scale, with weights only placed on one side. I need to figure out what the weights of these four pieces are.Okay, let's start by understanding the problem better. We have a total weight of 15 pounds that breaks into four integer weights. The key here is that these four pieces should be able to measure any weight from 1 to 15 pounds. That means, for every integer weight between 1 and 15, I should be able to select some combination of these four pieces to sum up to that weight.First, I think about how to represent numbers using sums of smaller numbers. This reminds me of binary numbers, where each digit represents a power of 2. For example, with weights of 1, 2, 4, and 8 pounds, you can combine them to make any number up to 15 pounds. Let me check that:- 1 = 1- 2 = 2- 3 = 1 + 2- 4 = 4- 5 = 1 + 4- 6 = 2 + 4- 7 = 1 + 2 + 4- 8 = 8- 9 = 1 + 8- 10 = 2 + 8- 11 = 1 + 2 + 8- 12 = 4 + 8- 13 = 1 + 4 + 8- 14 = 2 + 4 + 8- 15 = 1 + 2 + 4 + 8Wow, that works perfectly! Each number from 1 to 15 can be made by adding some combination of 1, 2, 4, and 8. So, if the four pieces are 1, 2, 4, and 8 pounds, they can measure any weight from 1 to 15 pounds.But wait, does this add up to 15 pounds in total? Let me check:1 + 2 + 4 + 8 = 15Yes, it does. So that fits the total weight requirement.Is there any other combination that could work? Maybe, but this seems like the most straightforward one. Let me think if there's another way to break down 15 into four integers that can cover all weights from 1 to 15.Suppose I try using different numbers. For example, what if one of the weights is 3 pounds? Then, can I still measure all weights up to 15? Let's see:If I have weights like 1, 3, 5, and 6, does that work? Let me try:- 1 = 1- 2 = ? Hmm, I don't have a 2. I can't make 2 with 1, 3, 5, 6.- So that doesn't work.What about 1, 2, 3, and 9? Let's check:- 1 = 1- 2 = 2- 3 = 3- 4 = 1 + 3- 5 = 2 + 3- 6 = 3 + 3, but I only have one 3.- So I can't make 6 with 1, 2, 3, 9.Hmm, that doesn't work either.What if I try 1, 2, 5, and 7?- 1 = 1- 2 = 2- 3 = 1 + 2- 4 = ? I don't have a 4. Can't make 4 with 1, 2, 5, 7.- So that doesn't work.It seems like using powers of 2 is the most efficient way to cover all the numbers without missing any. Each weight is double the previous one, which allows for binary combinations.Another thought: what if I use 1, 2, 3, and 9? Wait, I tried that earlier and it didn't work for 6. Maybe if I adjust differently.Alternatively, what if I use 1, 2, 4, and 8? That's the same as the binary approach, and it works perfectly.Is there a way to have non-powers of 2 and still cover all weights? Maybe, but it's more complicated. For example, using 1, 3, 4, and 7:- 1 = 1- 2 = ? Can't make 2.- So that's no good.Or 1, 2, 3, and 9:- 1 = 1- 2 = 2- 3 = 3- 4 = 1 + 3- 5 = 2 + 3- 6 = 3 + 3, but only one 3.- So again, can't make 6.It seems challenging to find another set that works as smoothly as the powers of 2. The key with powers of 2 is that each weight fills a unique "slot" in the binary representation, allowing every number to be expressed uniquely as a sum of these weights.Let me think about the balance scale. Since we can only place weights on one side, we don't have the option to subtract weights by placing them on the opposite side. So, we need to rely solely on addition to reach the desired weights.Given that, the binary approach is perfect because it allows us to build up any number through addition without gaps.Another consideration: the total weight must be 15. The sum of 1, 2, 4, and 8 is indeed 15, so that fits perfectly.Is there any other set of four integers that sum to 15 and can measure all weights from 1 to 15? I'm trying to think of another combination, but every time I try, I end up missing some numbers.For example, if I try 1, 2, 5, and 7:- 1 = 1- 2 = 2- 3 = 1 + 2- 4 = ? Can't make 4.- So that's no good.Or 1, 3, 5, and 6:- 1 = 1- 2 = ? Can't make 2.- So that's no good.It seems like the only way to cover all numbers without missing any is to use the binary approach with powers of 2.Wait, what if I use 1, 2, 3, and 9? Let's see:- 1 = 1- 2 = 2- 3 = 3- 4 = 1 + 3- 5 = 2 + 3- 6 = 3 + 3, but only one 3.- So can't make 6.Nope, still missing.What about 1, 2, 4, and 8? That works perfectly, as we saw earlier.I think I've convinced myself that the only way to have four integer weights that sum to 15 and can measure every integer weight from 1 to 15 is to use the powers of 2: 1, 2, 4, and 8.Just to double-check, let's list out all the combinations:- 1- 2- 1 + 2 = 3- 4- 1 + 4 = 5- 2 + 4 = 6- 1 + 2 + 4 = 7- 8- 1 + 8 = 9- 2 + 8 = 10- 1 + 2 + 8 = 11- 4 + 8 = 12- 1 + 4 + 8 = 13- 2 + 4 + 8 = 14- 1 + 2 + 4 + 8 = 15Yes, that covers every number from 1 to 15. So, the weights must be 1, 2, 4, and 8 pounds.

answer：Okay, so I need to find the coefficient of ( x^6 ) in the expansion of ( (x + 2)^8 ). Hmm, I remember something about the binomial theorem from algebra class. Let me try to recall how that works.The binomial theorem says that ( (a + b)^n ) can be expanded as the sum from ( k = 0 ) to ( n ) of ( binom{n}{k} a^{n - k} b^k ). So, in this case, ( a ) is ( x ), ( b ) is 2, and ( n ) is 8. That means each term in the expansion will look like ( binom{8}{k} x^{8 - k} cdot 2^k ).But I need the term where the exponent of ( x ) is 6. So, I need to find the value of ( k ) such that ( 8 - k = 6 ). Let me solve that: ( 8 - k = 6 ) implies ( k = 2 ). Okay, so the term with ( x^6 ) corresponds to ( k = 2 ).Now, I need to calculate the binomial coefficient ( binom{8}{2} ). I think the formula for the binomial coefficient is ( binom{n}{k} = frac{n!}{k!(n - k)!} ). Plugging in the numbers, ( binom{8}{2} = frac{8!}{2!(8 - 2)!} = frac{8!}{2!6!} ).Calculating that, ( 8! ) is ( 8 times 7 times 6! ), so it simplifies to ( frac{8 times 7 times 6!}{2 times 1 times 6!} ). The ( 6! ) cancels out from numerator and denominator, leaving ( frac{8 times 7}{2 times 1} = frac{56}{2} = 28 ). So, ( binom{8}{2} = 28 ).Next, I need to calculate ( 2^k ) where ( k = 2 ). That's straightforward: ( 2^2 = 4 ).Now, to find the coefficient of ( x^6 ), I multiply the binomial coefficient by ( 2^k ). So, ( 28 times 4 = 112 ).Wait, let me double-check that. If I have ( binom{8}{2} = 28 ) and ( 2^2 = 4 ), then ( 28 times 4 ) should indeed be 112. Yeah, that seems right.Just to be thorough, maybe I can think about another way to approach this. I know that the coefficients in the binomial expansion are symmetric, so ( binom{8}{k} = binom{8}{8 - k} ). That means ( binom{8}{2} = binom{8}{6} ), which is consistent with what I found earlier.Also, if I were to write out the expansion, the term with ( x^6 ) would be the third term (since ( k ) starts at 0). Let me list out the first few terms to see if that makes sense:- ( k = 0 ): ( binom{8}{0} x^8 cdot 2^0 = 1 cdot x^8 cdot 1 = x^8 )- ( k = 1 ): ( binom{8}{1} x^7 cdot 2^1 = 8 cdot x^7 cdot 2 = 16x^7 )- ( k = 2 ): ( binom{8}{2} x^6 cdot 2^2 = 28 cdot x^6 cdot 4 = 112x^6 )Yep, that matches what I calculated earlier. So, the coefficient of ( x^6 ) is indeed 112.I think I've got it right. The key steps were identifying the correct term using the binomial theorem, calculating the binomial coefficient, and then multiplying by the appropriate power of 2. Everything checks out, so I'm confident in the answer.