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B) Ta có: 2x-2y-x2+2xy-y2
⇔ 2(x-y)-(x2-2xy+y2)
⇔ 2(x-y)-(x-y)2
⇔ (x-y)(2-x+y)
Đúng thì tick nhé
\(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\dfrac{\left(x+y\right)\left(x+2y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}\)
\(=\dfrac{x+y}{x^2-y^2}\)
\(=\dfrac{1}{x-y}\)
1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)
\(B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}\)
\(B=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}\)
\(B=\dfrac{\left(x+2y\right)\left(x^2-y^2\right)}{x\left(x+y\right)+2y\left(x+y\right)}\)
\(B=\dfrac{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(2y+x\right)}\)
\(B=x-y\)\(\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)\)
Tham khảo nhé~
\(B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}\)
\(=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}\)
\(=\dfrac{\left(x^2-y^2\right)\left(x+2y\right)}{x\left(x+y\right)+2y\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x+y\right)}\)
\(=x-y\)
a/ \(\dfrac{x^2-x-6}{x^2+7x+10}=\dfrac{x^2-3x+2x-6}{x^2+2x+5x+10}=\dfrac{x\left(x-3\right)+2\left(x-3\right)}{x\left(x+2\right)+5\left(x+2\right)}=\dfrac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}=\dfrac{x-3}{x+5}\)
b/ \(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}=\dfrac{x^2+xy+2xy+2y^2}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}=\dfrac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x+2y\right)\left(x^2-y^2\right)}=\dfrac{\left(x+2y\right)\left(x+y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)
a) \(\dfrac{x^2-x-6}{x^2+7x+10}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x+5\right)}=\dfrac{x-3}{x+5}\)
b) \(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}=\dfrac{\left(x+y\right)\left(x+2y\right)}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}=\dfrac{x+y}{x^2-y^2}=\dfrac{x+y}{\left(x+y\right)\left(x-y\right)}=\dfrac{1}{x-y}\)