Đáp án:
a) x=0
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {5;8} \right\}\\
\dfrac{6}{{x - 5}} + \dfrac{{x + 2}}{{x - 8}} = \dfrac{{18}}{{\left( {x - 5} \right)\left( {8 - x} \right)}} - 1\\
\to \dfrac{{6\left( {x - 8} \right) + \left( {x + 2} \right)\left( {x - 5} \right) + 18 + \left( {x - 5} \right)\left( {x - 8} \right)}}{{\left( {x - 5} \right)\left( {x - 8} \right)}} = 0\\
\to 6x - 48 + {x^2} - 3x - 10 + 18 + {x^2} - 13x + 40 = 0\\
\to 2{x^2} - 10x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 5\left( l \right)
\end{array} \right.\\
b)\dfrac{{x - 4}}{{x + 1}} + \dfrac{{x + 4}}{{x + 1}}\\
= \dfrac{{x - 4 + x + 4}}{{x + 1}} = \dfrac{{2x}}{{x + 1}}
\end{array}\)
