Ta có:

\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)

\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)

\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)

\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)

\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)

b/ Thế vô rồi tính nhé

Đoạn gần cuối thay y-x= 1 luôn 

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)

\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\)  giờ mới thay không biết đã tối giản chưa

1. Có 1 thừa số là \(1-\frac{5}{5}=0\) nên tích sẽ bằng 0

2.\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{49}{50}=\frac{1}{50}\)

\(\frac{\left(-\frac{1}{3}\right)^2-\left(\frac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}=\frac{\frac{1}{9}-\frac{27}{64}.4}{-2+\frac{9}{16}-\frac{3}{8}}=\frac{\frac{1}{9}-\frac{27}{16}}{-2+\frac{3}{16}}\)

\(=\frac{\frac{16}{144}-\frac{243}{144}}{-\frac{32}{16}+\frac{3}{16}}=\frac{\frac{-227}{144}}{\frac{-29}{16}}=\frac{-227}{144}.\frac{-16}{29}\)

\(=\frac{227.16}{144.29}=\frac{227.1}{9.29}=\frac{227}{261}\)

Đáp số: \(\frac{227}{261}\)

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nhung hdt dang nho do ban