Đáp án:
$\begin{array}{l}
a)3x\left| {x + 1} \right| - 2x\left| {x + 2} \right| = 12\\
+ Khi:x \ge - 1\\
\Leftrightarrow 3x\left( {x + 1} \right) - 2x\left( {x + 2} \right) = 12\\
\Leftrightarrow 3{x^2} + 3x - 2{x^2} - 4x - 12 = 0\\
\Leftrightarrow {x^2} - x - 12 = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow x = 4\left( {do:x \ge - 1} \right)\\
+ Khi: - 2 \le x < - 1\\
3x\left( { - x - 1} \right) - 2x\left( {x + 2} \right) = 12\\
\Leftrightarrow - 3{x^2} - 3x - 2{x^2} - 4x = 12\\
\Leftrightarrow 5{x^2} + 7x + 12 = 0\left( {vn} \right)\\
+ Khi:x < - 2\\
\Leftrightarrow 3x\left( { - x - 1} \right) + 2x\left( {x + 2} \right) = 12\\
\Leftrightarrow - 3{x^2} - 3x + 2{x^2} + 4x = 12\\
\Leftrightarrow {x^2} - x + 12 = 0\left( {vn} \right)\\
\text{Vậy}\,x = 4\\
b)\dfrac{{{x^2} - 4 - \left| {x + 1} \right|}}{2} = x\left( {x + 1} \right)\\
\Leftrightarrow {x^2} - 4 - \left| {x + 1} \right| = 2x\left( {x + 1} \right)\\
+ Khi:x \ge - 1\\
\Leftrightarrow {x^2} - 4 - x - 1 = 2{x^2} + 2x\\
\Leftrightarrow {x^2} + 3x + 5 = 0\left( {vn} \right)\\
+ Khi:x < - 1\\
\Leftrightarrow {x^2} - 4 + x + 1 = 2{x^2} + 2x\\
\Leftrightarrow {x^2} + x + 3 = 0\left( {vn} \right)\\
\text{Vậy phương trình vô nghiệm}\\
c)Dkxd:x\# 0;x\# 2;x\# 16\\
\dfrac{7}{{8x}} - \dfrac{{x - 1}}{{2x\left( {x - 2} \right)}} - \dfrac{1}{{x - 16}} = \dfrac{{x - 5}}{{4{x^2} - 8x}}\\
\Leftrightarrow \dfrac{{7\left( {x - 2} \right) - 4\left( {x - 1} \right)}}{{8x\left( {x - 2} \right)}} - \dfrac{{x - 5}}{{4x\left( {x - 2} \right)}} = \dfrac{1}{{x - 16}}\\
\Leftrightarrow \dfrac{{3x - 10 - 2x + 10}}{{8x\left( {x - 2} \right)}} = \dfrac{1}{{x - 16}}\\
\Leftrightarrow \dfrac{x}{{8x\left( {x - 2} \right)}} = \dfrac{1}{{x - 16}}\\
\Leftrightarrow \dfrac{1}{{8\left( {x - 2} \right)}} = \dfrac{1}{{x - 16}}\\
\Leftrightarrow 8x - 16 = x - 16\\
\Leftrightarrow x = 0\left( {ktm} \right)\\
\text{Vậy phương trình vô nghiệm}\\
d)Dkxd:x\# 3;x\# - 3\\
\dfrac{{{x^2} - x}}{{x + 3}} - \dfrac{{{x^2}}}{{x - 3}} = \dfrac{{7{x^2} - 3x}}{{9 - {x^2}}}\\
\Leftrightarrow \dfrac{{\left( {{x^2} - x} \right)\left( {x - 3} \right) - {x^2}\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
= \dfrac{{ - 7{x^2} + 3x}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\Leftrightarrow {x^3} - 4{x^2} + 3x - {x^3} - 3{x^2} = - 7{x^2} + 3x\\
\Leftrightarrow 0 = 0\left( {tm} \right)
\end{array}$
Vậy pt nghiệm đúng với mọi x#3; x#-3
