Với đk trên ta có:

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

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

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

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

\(=\frac{2}{x}+\frac{x-y}{xy}\)

\(=\frac{x+y}{xy}\)

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

\(A=\left[\frac{x^2-y^2}{xy}-\left(\frac{x}{y^2}-\frac{y}{x^2}\right)\right].\frac{xy}{x-y}\) => \(A=\left(\frac{x^2-y^2}{xy}-\frac{x^3-y^3}{x^2y^2}\right).\frac{xy}{x-y}=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^2y^2}\right).\frac{xy}{x-y}\)

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

Làm ơn giúp em bài toán này . 

ĐKXĐ : \(x,y\ne0\)\(;\)\(x\ne y\)

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

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

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

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

\(P=\frac{2y}{xy}-\frac{x+y}{xy}=\frac{y-x}{xy}\)

\(b)\)

+) Với \(\left|2x-1\right|=1\)\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

Mà \(x\ne0\) ( ĐKXĐ ) nên \(x=1\)

+) Với \(\left|y+1\right|=\frac{1}{2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y+1=\frac{1}{2}\\y+1=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{-1}{2}\\y=\frac{-3}{2}\end{cases}}}\)

Thay \(x=1;y=\frac{-1}{2}\) vào \(A=\frac{y-x}{xy}\) ta được : \(A=\frac{\frac{-1}{2}-1}{1.\frac{-1}{2}}=\frac{\frac{-3}{2}}{\frac{-1}{2}}=3\)

Thay \(x=1;y=\frac{-3}{2}\) vào \(A=\frac{y-x}{xy}\) ta được : \(A=\frac{\frac{-3}{2}-1}{1.\frac{-3}{2}}=\frac{\frac{-5}{2}}{\frac{-3}{2}}=\frac{15}{4}\)

Vậy ... 

Cảm ơn nè <3 

chịu thôi không biết đâu

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

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

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

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

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

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

Giờ chỉ cần thế x, y vô nữa là xong nhé.

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

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

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

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

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

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

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

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

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

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

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

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

Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :

\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)

Ừ nhở chị sai từ chỗ \(\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2}{y\left(x+y\right)}=\frac{y^2+2xy}{y\left(x+y\right)}\)em nhé 

\(a,ĐKXĐ:x\ne-;y\ne0\)

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

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

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

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

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

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

\(P=\frac{2}{x}-\frac{x-y}{xy}=\frac{2y-x+y}{xy}=\frac{3y-x}{xy}\)

\(b,x^2+y^2+10=2\left(x-3y\right)\)

\(\Leftrightarrow x^2+y^2+10=2x-6y\)

\(\Leftrightarrow x^2-2x+1+y^2+6y+9=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

thay vào P được : \(P=\frac{3\left(-3\right)-1}{-3\cdot1}=\frac{-10}{-3}=\frac{10}{3}\)

a, Rút gọn A

b,Tìm giá trị P, biết x,y thỏa mãn đẳng thức

x^2+y^2+10=2(x-3y)

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

ĐK: \(\hept{\begin{cases}x,y\ne0\\x\ne-y\end{cases}}\)

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

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

\(A=\frac{2}{x}+\frac{x-y}{xy}\)

\(A=\frac{2y+x-y}{xy}\)

\(A=\frac{x+y}{xy}\)