Đáp án:
$\begin{array}{l}
d)Dkxd:x \ne 2;x \ne - 2\\
\dfrac{{2x - 3}}{{x + 2}} - \dfrac{{x + 2}}{{x - 2}} = \dfrac{2}{{{x^2} - 4}}\\
\Leftrightarrow \dfrac{{\left( {2x - 3} \right).\left( {x - 2} \right) - \left( {x + 2} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{2}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\Leftrightarrow 2{x^2} - 4x - 3x + 6 - \left( {{x^2} + 4x + 4} \right) = 2\\
\Leftrightarrow {x^2} - 11x + 2 = 2\\
\Leftrightarrow {x^2} - 11x = 0\\
\Leftrightarrow x\left( {x - 11} \right) = 0\\
\Leftrightarrow x = 0;x = 11\left( {tmdk} \right)\\
Vậy\,x = 0;x = 11\\
B6)\\
a)\left| {2x - 3} \right| = 4\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 4\\
2x - 3 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{2};x = - \dfrac{1}{2}\\
b)\left| {3x - 1} \right| - x = 2\\
\Leftrightarrow \left| {3x - 1} \right| = x + 2\left( {dk:x \ge - 2} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = x + 2\\
3x - 1 = - x - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\left( {tm} \right)\\
x = - \dfrac{1}{4}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{3}{2};x = - \dfrac{1}{4}\\
c)Dkxd:x \ge - \dfrac{3}{2}\\
\left| {x - 7} \right| = 2x + 3\\
\Leftrightarrow \left[ \begin{array}{l}
x - 7 = 2x + 3\\
x - 7 = - 2x - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 10\left( {ktm} \right)\\
x = \dfrac{4}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{4}{3}\\
d)\left| {x - 4} \right| + 3x = 5\\
\Leftrightarrow \left| {x - 4} \right| = 5 - 3x\left( {dk:x \le \dfrac{5}{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = 5 - 3x\\
x - 4 = 3x - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{9}{4}\left( {ktm} \right)\\
x = \dfrac{1}{2}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2}\\
e)2\left( {x + 1} \right)\left| {x - 4} \right| = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 4
\end{array} \right.\\
Vậy\,x = - 1;x = 4\\
j)\left| {x - 7} \right| = 2\\
\Leftrightarrow \left[ \begin{array}{l}
x - 7 = 2\\
x - 7 = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x = 5
\end{array} \right.\\
Vậy\,x = 9;x = 5\\
l)\left| {5 - 2x} \right| = 1 - x\left( {dk:x \le 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
5 - 2x = 1 - x\\
5 - 2x = x - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\left( {ktm} \right)\\
x = 2\left( {ktm} \right)
\end{array} \right.
\end{array}$
Vậy pt vô nghiệm
