ĐKXĐ: x²-y²\(\ne\)0                                                      4xy\(\ne\)0

     \(\Leftrightarrow\)\(\left(x-y\right)\left(x+y\right)\ne0\)                            <=>x\(\ne\)0 và y \(\ne\)0

     \(\Leftrightarrow x\ne y\) và \(x\ne-y\)

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

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

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

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

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

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..

ĐKXĐ : \(x\ne\pm y\)

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

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

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

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

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

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

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

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

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!

mơn bn nhìu na!!!

tks pạn nhìu nhoa

hay đó, cảm ơn luôn nha!~~ (dù ko lq :D)

Lời giải:

ĐK: $x\neq 5;x\neq 0; y\neq 2; y\neq -1$

\(M=\frac{x^2-25}{x^3-10x^2+25x}:\frac{y-2}{(y-2)(y+1)}=\frac{(x-5)(x+5)}{x(x^2-10x+25)}:\frac{1}{y+1}\)

\(=\frac{(x-5)(x+5)}{x(x-5)^2}:\frac{1}{y+1}=\frac{x+5}{x(x-5)}.(y+1)=\frac{(x+5)(y+1)}{x(x-5)}\)

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$x^2+9y^2-4xy=2xy-|x-3|$

$\Leftrightarrow x^2+9y^2-6xy=-|x-3|$

$\Leftrightarrow (x-3y)^2+|x-3|=0$

Dễ thấy $(x-3y)^2\geq 0; |x-3|\geq 0$ với mọi $x,y\in $ĐKXĐ nên để tổng của chúng bằng $0$ thì:

$x-3y=x-3=0\Rightarrow x=3; y=1$

Khi đó: $M=\frac{(3+5)(1+1)}{3(3-5)}=\frac{-8}{3}$