Đáp án:
d) x=11
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} + 6x + 9 - {x^2} + 6x - 9 = 6x + 18\\
\to 6x = 18\\
\to x = 3\\
b)DK:x \ne \pm 5\\
\dfrac{3}{{2\left( {x + 5} \right)}} - \dfrac{{2x}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} + \dfrac{3}{{x - 5}} = 0\\
\to \dfrac{{3\left( {x - 5} \right) - 4x + 3\left( {x + 5} \right)}}{{2\left( {x - 5} \right)\left( {x + 5} \right)}} = 0\\
\to 3x - 15 - 4x + 3x + 15 = 0\\
\to 2x = 0 \to x = 0\\
d)DK:x \ne \left\{ {1;3} \right\}\\
\dfrac{{x + 5}}{{x - 1}} - \dfrac{{x + 1}}{{x - 3}} = \dfrac{8}{{{x^2} - 4x + 3}}\\
\to \dfrac{{\left( {x + 5} \right)\left( {x - 3} \right) - \left( {x + 1} \right)\left( {x - 1} \right) - 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \dfrac{{{x^2} + 2x - 15 - {x^2} + 1 - 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to 2x - 22 = 0\\
\to x = 11
\end{array}\)
( bạn xem lại đề câu c nha )
