Question

Giải pt: \(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}-1=\dfrac{3}{x^2-x-2}\)

Answer 1

`ĐKXĐ: x ne -1 ; x ne 2`

\(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}-1=\dfrac{3}{x^2-x-2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\dfrac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\dfrac{3}{\left(x-2\right)\left(x+1\right)}=1\\ \Leftrightarrow\dfrac{x^2-4+3x+3-3}{x^2-x-2}=1\\ \Rightarrow x^2+3x-4=x^2-x-2\\ \Leftrightarrow4x-2=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)