a)\(M=\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\left(ĐKXĐ:x\ne-1;y\ne1\right)\)

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

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

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

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

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

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

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

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

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

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

             \(M=x+xy-y\)

b)Ta có:\(x+xy-y=-7\)

            \(x\left(y+1\right)-y-1+8=0\)

             \(\left(x-1\right)\left(y+1\right)=-8\)

Ta có : -8 = 8 . -1 = -8 . 1 = -2.4=-4.2

       Rồi chỗ đó tự thay nha

Đây là bài dài nhất trong olm của mk

a.\(P=x^2-y^2+x^3+y^3-x^3y^2-x^2y^3\) phần (x+y)(1-y)(1+x)

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

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

b/Dễ r

\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)

\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)

\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)

\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)

Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)

\(\Leftrightarrow A\ne0\forall x;y\)

tra loi nhanh di ae

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

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

kết quả là -1 nha!