Ta có: \(\dfrac{x^2+xy}{x^2+xy+y^2}-\left(\dfrac{x\left(2x^2+xy-y^2\right)}{x^3-y^3}-2+\dfrac{y}{y-x}\right):\dfrac{x-y}{x}-\dfrac{x}{x-y}\)

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

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

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

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

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

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

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

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

Ta có: \(x^2-2y^2=xy\)

\(\Leftrightarrow x^2-xy-2y^2=0\)

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

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

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

Vì \(x+y\ne0\) nên x-2y=0

hay x=2y

Thay x=2y vào biểu thức \(A=\dfrac{x-y}{x+y}\), ta được: 

\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

Vậy: \(A=\dfrac{1}{3}\)

ĐKXĐ: \(...\)

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

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

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

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

\(P=\dfrac{2}{x}+\dfrac{x-y}{xy}=\dfrac{2}{x}+\dfrac{1}{y}-\dfrac{1}{x}=\dfrac{1}{x}+\dfrac{1}{y}\)

b/ \(x^2+y^2+10=2x-6y\Leftrightarrow x^2-2x+1+y^2+6y+9=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+3\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{3}\)

a/ \(B=\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\left(\dfrac{x+y}{x^2+xy+y^2}+\dfrac{1}{x-y}\right)\)

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

\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{x^2-y^2+x^2+xy+y^2}{x^3-y^3}\)

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

b/ Khi x = -1/2 và y = 3 ta có:

\(B=\dfrac{2\cdot\left(-\dfrac{1}{2}\right)+3}{3}=\dfrac{-1+3}{3}=\dfrac{2}{3}\)

Tử Đằng Sao t cứ có cảm giác m đang tự lừa mình dối người thế -.- đừng có nói là m k biết làm bài này nhé

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

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

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

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

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

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

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

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

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

SỬa đề: x^3-xy^2

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

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

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

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

Để A>0 thì y>0