Question

[ngọc ánh 2k8]

Giải các phương trình 
\(\dfrac{x+2}{x-1}\)- \(\dfrac{x-2}{x+1}\) =4

Answer 1

[Hà Quang Minh]

ĐKXĐ: \(x\ne1;x\ne-1\)

\(\dfrac{x+2}{x-1}-\dfrac{x-2}{x+1}=4\\ \)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+1\right)-\left(x-2\right)\left(x-1\right)}{x^2-1}=4\\\)

\( \Leftrightarrow\dfrac{x^2+3x+2-x^2+3x-2}{x^2-1}=4\\ \)

\(\Leftrightarrow6x=4x^2-4\\ \Leftrightarrow2x^2-3x-2=0\\ \)

\(\Leftrightarrow\left(2x+1\right)\left(x-2\right)=0\\ \)

\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\left(tm\right)\)

Vậy \(x\in\left\{-\dfrac{1}{2};2\right\}\)