Tập nghiệm của phương trình sau \(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3 x^{2}-12}\) là

* Hướng dẫn giải

\(\begin{array}{l} \frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3 x^{2}-12} \\ \Leftrightarrow \frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6}{3(x-2)(x+2)} \end{array}\)

ĐK: \(\left\{\begin{array}{l} x-2 \neq 0 \\ x+2 \neq 0 \end{array} \Leftrightarrow\left\{\begin{array}{l} x \neq 2 \\ x \neq-2 \end{array}\right.\right.\)

Khi đó

\(\begin{array}{l} \frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3 x^{2}-12} \\ \Leftrightarrow \frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3(x-2)(x+2)} \\ \Leftrightarrow \frac{1}{x+2}-\frac{6-x}{3(x-2)(x+2)}+\frac{1}{x-2}-1=0 \\ \Leftrightarrow \frac{3(x-2)}{3(x-2)(x+2)}-\frac{6-x}{3(x-2)(x+2)}+\frac{3(x+2)}{3(x-2)(x+2)}-\frac{3(x-2)(x+2)}{3(x-2)(x+2)}=0 \\ \Leftrightarrow 3 x-6-6+x+3 x+6-3 x^{2}+12=0 \\ \Leftrightarrow 3 x^{2}-7 x-6=0 \\ \Leftrightarrow\left[\begin{array}{l} x=\frac{-2}{3}(n) \\ x=3(n) \end{array}\right. \end{array}\)

Vậy \(S=\left\{-\frac{2}{3} ; 3\right\}\)