Question

giải pt

\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)

Answer 1

ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2-3x+2-x^2-2x-2+5x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}

Answer 2

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)