Question

Giải pt: 3/x²+x-2 - 1/x-1 = -7/x+2

Answer 1

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

<=> \(\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x-1}=\frac{-7}{x+2}\)

<=> \(\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\cdot\left(x+2\right)}=\frac{-7.\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\)

=> \(3-x-2=-7x+7\)

<=> \(3-2-7=x-7x\)

<=> \(-6=-6x\)

<=> \(x=1\)

Vậy phương trình có nghiệm x = 1