Câu hỏi của Mina

Question

-x2 + 2xy - y2 / x+ y = ? / y2 - x2

ILoveMath · Accepted Answer

$\dfrac{-x^2+2xy-y^2}{x+y}=\dfrac{-\left(x^2-2xy+y^2\right)}{x+y}=\dfrac{-\left(x-y\right)^2}{\left(x+y\right)}=\dfrac{-\left(x-y\right)^3}{\left(x+y\right)\left(x-y\right)}=\dfrac{-\left(x-y\right)^3}{x^2-y^2}=\dfrac{\left(x-y\right)^3}{y^2-x^2}\Rightarrow?=\left(x-y\right)^3$Câu trả lời: $\dfrac{-x^2+2xy-y^2}{x+y}=\dfrac{-\left(x^2-2xy+y^2\right)}{x+y}=\dfrac{-\left(x-y\right)^2}{\left(x+y\right)}=\dfrac{-\left(x-y\right)^3}{\left(x+y\right)\left(x-y\right)}=\dfrac{-\left(x-y\right)^3}{x^2-y^2}=\dfrac{\left(x-y\right)^3}{y^2-x^2}\Rightarrow?=\left(x-y\right)^3$

Minh Hiếu · Answer

$\dfrac{-x^2+2xy-y^2}{x+y}=\dfrac{?}{y^2-x^2}$$\dfrac{-\left(x-y\right)^2}{x+y}=\dfrac{-?}{x^2-y^2}$$\dfrac{-\left(x-y\right)^2}{x+y}=\dfrac{-?}{\left(x-y\right)\left(x+y\right)}$$-?\left(x+y\right)=-\left(x-y\right)^3\left(x+y\right)$$?=\left(x-y\right)^3$Câu trả lời: $\dfrac{-x^2+2xy-y^2}{x+y}=\dfrac{?}{y^2-x^2}$$\dfrac{-\left(x-y\right)^2}{x+y}=\dfrac{-?}{x^2-y^2}$$\dfrac{-\left(x-y\right)^2}{x+y}=\dfrac{-?}{\left(x-y\right)\left(x+y\right)}$$-?\left(x+y\right)=-\left(x-y\right)^3\left(x+y\right)$$?=\left(x-y\right)^3$

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