Question

Rút gọn: \(\left(\frac{1}{x^2-xy}-\frac{3y^2}{x^4-xy^3}-\frac{y}{x^3+x^2y+xy^2}\right):\frac{x+y}{x^2+xy+y^2}\)

Answer 1

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

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

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

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

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

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

Answer 2

Mình vs bạn trùng họ và tên rồi thì phải....!