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

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

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

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

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

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

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

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

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

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

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

ib tui làm cho 

Câu a) bạn Despacito làm sai kq r. Kq dúng là A=2x(x+y).

Câu b)

\(3x^2+y^2+2x-2y-1=0\)

\(\Leftrightarrow2x^2+2xy+x^2-2xy+y^2+2x-2y-1=0\)

\(\Leftrightarrow2x\left(x+y\right)+\left(x-y\right)^2+2\left(x-y\right)+1-2=0\)

\(\Leftrightarrow2A+\left(x-y+1\right)^2-2=0\)

\(\Leftrightarrow\left(x-y+1\right)^2=0\)

\(\Leftrightarrow x-y+1=0\)

\(\Leftrightarrow x-y=-1\)

Câu a bạn rút gọn A đc bao nhiêu

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

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

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

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

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

\(A=x\left(x+y\right)\)

\(A=\frac{4xy}{y^2-x^2}:\left(\frac{1}{y^2+2xy+x^2}-\frac{x^3+y^3}{x^4-y^4}\right)\left(x\ne\pm y;y\ne0\right)\)

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

(a) làm được rồi port lên luôn vì (b) cần cái KQ của (a)

Rút gọn ra \(A=y+x\) nhé

có vẻ A không gọn thế 

\(P=2x\left(x+y\right)=2x^2+2xy\) Với x khác y, x khác -y

\(3x^2+y^2+2x-2y=1\)\(\Leftrightarrow2x^2+2xy+y^2+x^2+1-2xy+2x-2y=2\)

\(\Leftrightarrow P+\left(x-y+1\right)^2=2\)\(\Leftrightarrow P=2-\left(x-y+1\right)^2\le2\)vì \(\left(x-y+1\right)^2\ge0\)với mọi x, y là số thực

Vì P nguyên dương => P=1 

Khi đó \(\left(x-y+1\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-y+1=-1\\x-y+1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=-2\\x-y=0\left(loai\right)\end{cases}}\)

vì x khác y