Question

Rút gọn phân thức:

\(a,\dfrac{x^2-x-6}{x^2+7x+10}\)

\(b,\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

Answer 1

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

Answer 2

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