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

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

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

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

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

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

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

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

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

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

      \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

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

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

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

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

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

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

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

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

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

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

mơn bn nhìu na!!!

tks pạn nhìu nhoa

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