Question

\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)

Answer 1

ĐK: \(x\ne1;x\ne-2\)

\(Pt\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}\)

\(\Rightarrow3-\left(x+2\right)=-7x+7\)

\(\Leftrightarrow x=1\) (ktm)
Vậy pt vô nghiệm

Answer 2

ĐK: `x \ne 1; -2`

`3/(x^2+x-2)-1/(x-1)=-7/(x+2)`

`<=> 3-(x+2)=-7(x-1)`

`<=> 3-x-2=-7x+7`

`<=> 6x = 6`

`<=> x=1 (L)`

Vậy `S={∅}`.