Ta có: \(A=\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)

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

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

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

\(=\frac{1}{\left(x+y\right).2x}\)

Kb với mình nha mn!

ĐKXĐ : \(x\ne\pm y\)

Ta có : \(A=\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)

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

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

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

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

=> \(A=\frac{1}{2x\left(x+y\right)}\)

ĐKXĐ : \(x\ne\mp y\) ; \(x,y\ne0\)

Ta có :

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

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

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

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

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

\(=\frac{1}{2x\left(x+y\right)}\)

Vậy..

(\(\frac{\left(x+y\right)^2}{x+y}\) -\(\frac{4xy}{x+y}\) ):\(\frac{\left(x-y\right)^2}{\left(x+y\right)\left(x-y\right)}\)

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

a) ĐKXĐ: \(x\ne\pm1\)

\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{1-x}{\left(1+x\right)\left(1-x\right)}-\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{x}{x^2-1}\right)\)

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

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

\(=\frac{1-4x}{-x^2-3x+1}=\frac{4x-1}{x^2+3x-1}\) (chắc hết rút gọn được rồi)

Ơ sao câu trả lời của mình có khung màu vàng nhỉ?

\(\frac{4xy}{y^2-x^2}:\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right)\)

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

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

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

\(\frac{4xy}{y^2-x^2}:\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right)\)

\(=\frac{1}{\left(y-x\right)\left(y+x\right)}:\left(\frac{1}{\left(x+y\right)}-\frac{1}{\left(x-y\right)\left(x+y\right)}\right)\)

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

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

ib tui làm cho