Question

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

 

Answer 1

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

Answer 2

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

Answer 3

xin lỗi mk nhầm cái đoạn \(\frac{-\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^3}\) phải là \(\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}\) ms đúg