xin lỗi, chỉ có 1 trg hợp thôi

hình như bạn chép sai đề thì phải

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

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

Ta có:

\(\left(y^2+y+1\right)\left(x^2+x+1\right)\)

\(=x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1\)

\(=x^2y^2+x^2+y^2+2xy+2=x^2y^2+3\)

Ta lại có:

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

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

Theo đề bài ta có: (sửa đề luôn)

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(=\frac{\left(y^2+y+1\right)-\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

Em xin đóng góp cách 2 ạ 

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}\)

\(=\frac{x^4-x-y^4+y}{x^3y^3-y^3-x^3+1}\)

\(=\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)-\left(x-y\right)}{x^3y^3-\left(x^3+y^3\right)+1}\)

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

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

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

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

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

Do \(\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)

\(gt\Rightarrow y-1=-x\Rightarrow x-1=-y\)

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Leftrightarrow\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}+\frac{2\left(x-y\right)}{x^2y^2-3}=0\)

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

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

\(\Leftrightarrow\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+2xy+x^2+y^2+2\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(dpcm\right)\)

<=> x+y+2=xy

<=> y+2=xy-x

<=> y+2=x(y-1)

<=> x= (y+2)/(y-1)=(y-1+3)/(y-1)= 1+ 3/(y-1)

Vậy, để x nguyên thì y-1 phải là ước của 3

=> y-1={-3; -1; 1; 3}

=> y={-2; 0; 2; 4}

=> x={0; -2; 4; 2}

Do x, y khác 0 nên các cặp x, y thỏa mãn là (4; 2) và (2; 4)

Áp dụng bđt côsi cho 2 số dương lần lượt ta có : 

\(1+\frac{y}{x}\ge2\sqrt{\frac{y}{x}}\)

\(1+\frac{z}{y}\ge2\sqrt{\frac{z}{y}}\)

\(1+\frac{x}{z}\ge2\sqrt{\frac{x}{z}}\)

Nhân vế theo vế ta đc : \(\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\ge8\sqrt{\frac{xyz}{xyz}}=8\)

Dấu  = xảy ra khi : \(1=\frac{y}{x}\)=> x=y  và \(1=\frac{z}{y}\) => z=y và \(1=\frac{x}{z}\) => x=z

=> x=y=z

Thay vào M ta được : \(M=\frac{x^2}{2x^2}+\frac{y^2}{2y^2}+\frac{z^2}{2z^2}=\frac{3}{2}\).