ta có : `x^2 -x-6=(x-3)(x+2)`
`x+2=x+2`
MTC : `(x-3)(x+2)`
NTP : `1;x-3`
quy đồng :
`1/(x^2-x-6)=(1.1)/((x^2-x-6)1)=1/((x-3)(x+2))`
`(x-5)/(x+2)=((x-5)(x-3))/((x+2)(x-3))= ((x-5)(x-3))/((x-3)(x+2))`
\(\dfrac{1}{x^2-x-6}=\dfrac{1}{\left(x-3\right)\left(x+2\right)}\)
\(\dfrac{x-5}{x+2}=\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+2\right)}\)