Đáp án: $x\in \left \{ -\dfrac{1}{2};5 \right \}$

Giải thích các bước giải:

$\dfrac{x^{2}-7x-6}{3x+1}+2=\dfrac{x^{2}-5x-6}{x+1}(ĐK:x\neq -\dfrac{1}{3};x\neq -1\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}+2=-\dfrac{x^{2}+x-6x-6}{x+1}\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}+2=-\dfrac{x(x+1)-6(x+1)}{x+1}\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}+2=-\dfrac{(x+1)(x-6)}{x+1}\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}+2=-(x-6)\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}+2=-x+6\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}+2+x-6=0\\\Leftrightarrow \dfrac{x^{2}-7x-6}{3x+1}-4+x=0\\\Leftrightarrow \dfrac{x^{2}-7x-6-4(3x+1)+x(3x+1)}{3x+1}=0\\\Leftrightarrow \dfrac{x^{2}-7x-6-12x-4+3x^{2}+x}{3x+1}=0\\\Leftrightarrow \dfrac{4x^{2}-18x-10}{3x+1}=0\\\Leftrightarrow 4x^{2}-18x-10=0\\\Leftrightarrow 2x^{2}-9x-5=0\\\Leftrightarrow 2x^{2}+x-10x-5=0\\\Leftrightarrow x(2x+1)-5(2x+1)=0\\\Leftrightarrow (2x+1)(x-5)=0\\\Leftrightarrow \left[ \begin{array}{l}2x+1\\x-5=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=5\end{array} \right.$
Vậy $x\in \left \{ -\dfrac{1}{2};5 \right \}$

Đáp án:

Giải thích các bước giải: