Question

Rút gọn

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

Answer 1

2y^2 +xy -x^2 =y(y+x) +y^2 -x^2 =(x+y)(2y-x)

4x^2 +4x^2 y +y^2 -4 =4x^2 (y+1) +y^2-4 có vẻ hệ số lệch lại nhỉ

x^2 +y +xy +x =x(x+y) +x+y =(x+y) (x+1)

\(B=\dfrac{x-y}{2y-x}+\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}=\dfrac{x^2-y^2+\left(x^2+y^2+y-2\right)}{\left(x+y\right)\left(2y-x\right)}=\dfrac{2x^2+y-2}{\left(x+y\right)\left(2y-x\right)}\)\(C=\dfrac{4x^2\left(y+1\right)+y^2-4}{\left(x+y\right)\left(x+1\right)}\)

\(A=B:C=\dfrac{2x^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\dfrac{\left(x+y\right)\left(x+1\right)}{4x^2\left(y+1\right)+y^2-4}\)

\(A=\dfrac{2x^2+y-2}{\left(2y-x\right)}.\dfrac{\left(x+1\right)}{4x^2\left(y+1\right)+y^2-4}\)