Tập nghiệm của \(\frac{x-1}{3-x}-\frac{2-9 x}{x^{2}-x-6}=1-\frac{2 x}{x+2}\) là

* Hướng dẫn giải

ĐK: \(\left\{\begin{array}{l} 3-x \neq 0 \\ x+2 \neq 0 \end{array} \Leftrightarrow\left\{\begin{array}{l} x \neq 3 \\ x \neq-12 \end{array}\right.\right.\)

Khi đó

\(\begin{array}{l} \frac{x-1}{3-x}-\frac{2-9 x}{x^{2}-x-6}=1-\frac{2 x}{x+2}\\ \Leftrightarrow \frac{-x+1}{x-3}-\frac{2-9 x}{(x-3)(x+2)}=1-\frac{2 x}{x+2} \\ \Leftrightarrow \frac{(-x+1)(x+2)}{(x-3)(x+2)}-\frac{2-9 x}{(x-3)(x+2)}=\frac{(x-3)(x+2)}{(x-3)(x+2)}-\frac{2 x(x-3)}{(x+2)(x-3)} \\ \Leftrightarrow(-x+1)(x+2)-(2-9 x)=(x-3)(x+2)-2 x(x-3) \\ \Leftrightarrow-x^{2}-x+2-2+9 x=x^{2}-x-6-2 x^{2}+6 x \\ \Leftrightarrow-x^{2}+8 x=-x^{2}+5 x-6 \\ \Leftrightarrow 3 x=-6 \\ \Leftrightarrow x=-2(l) \\ \rightarrow S=\varnothing \end{array}\)