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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
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
\(A=\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\frac{y^2-xz}{\left(y+z\right)\left(y+x\right)}+\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
\(=\frac{\left(x^2-yz\right)\left(y+z\right)+\left(y^2-xz\right)\left(z+x\right)+\left(z^2-xy\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{x^2y+x^2z-y^2z-yz^2+y^2z+y^2x-xz^2-x^2z+z^2x+z^2y-x^2y-xy^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{0}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)
Vậy : \(A=0\)
\(\frac{(x^2-yz)(y+z)}{(x+y)(x+z)(y+z)}\) = \(\frac{(y^2-xz)(x+z)}{(x+y)(x+z)(y+z)}\)= \(\frac{(z^2-xy)(x+y)}{(x+y)(x+z)(y+z)}\)
quy đồng mẫu thức ta được
\(\frac{yz\left(z-y\right)+xz\left(x-z\right)+xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)\(=\frac{yz\left(z-y\right)+xz\left(x-z\right)-xy\left[\left(z-y\right)+\left(x-z\right)\right]}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{y\left(z-y\right)\left(z-x\right)+x\left(x-z\right)\left(z-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(z-y\right)\left(z-x\right)\left(y-x\right)}{xyz\left(z-y\right)\left(z-x\right)\left(y-x\right)}=\frac{1}{xyz}\)