Đáp án:
$\begin{array}{l}
d){\left( {x + 2} \right)^2} - 2\left( {x - 3} \right) = {\left( {x + 1} \right)^2}\\
\Rightarrow {x^2} + 4x + 4 - 2x + 6 - {x^2} - 2x - 1 = 0\\
\Rightarrow 9 = 0\left( {vl} \right)\\
\text{Vậy pt vô nghiệm}\\
e){\left( {x - 1} \right)^2} + {\left( {x - 2} \right)^2} = 2{\left( {x + 4} \right)^2} - \left( {22x + 27} \right)\\
\Rightarrow {x^2} - 2x + 1 + {x^2} - 4x + 4\\
= 2{x^2} + 16x + 32 - 22x - 27\\
\Rightarrow 5 = 5\left( {tm} \right)\\
\text{Vậy pt nghiệm đúng với mọi x}\\
g)\dfrac{{{{\left( {x - 1} \right)}^2}}}{3} + \dfrac{{{{\left( {x + 3} \right)}^2}}}{6} = \dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{2}\\
\Rightarrow \dfrac{{2\left( {{x^2} - 2x + 1} \right)}}{6} + \dfrac{{{x^2} + 6x + 9}}{6}\\
= \dfrac{{3\left( {{x^2} - x - 2} \right)}}{6}\\
\Rightarrow 2{x^2} - 4x + 2 + {x^2} + 6x + 9 = 3{x^2} - 3x - 6\\
\Rightarrow 5x = - 17\\
\Rightarrow x = \dfrac{{ - 17}}{5}\\
\text{Vậy}\,x = \dfrac{{ - 17}}{5}\\
h)\dfrac{{392 - x}}{{32}} + \dfrac{{390 - x}}{{34}} + \dfrac{{388 - x}}{{36}}\\
+ \dfrac{{386 - x}}{{38}} + \dfrac{{384 - x}}{{40}} = - 5\\
\Rightarrow \dfrac{{392 - x}}{{32}} + 1 + \dfrac{{390 - x}}{{34}} + 1 + \dfrac{{388 - x}}{{36}} + 1\\
+ \dfrac{{386 - x}}{{38}} + 1 + \dfrac{{384 - x}}{{40}} + 1 = 0\\
\Rightarrow \dfrac{{424 - x}}{{32}} + \dfrac{{424 - x}}{{34}} + \dfrac{{424 - x}}{{36}}\\
+ \dfrac{{424 - x}}{{38}} + \dfrac{{424 - x}}{{40}} = 0\\
\Rightarrow \left( {424 - x} \right).\left( {\dfrac{1}{{32}} + \dfrac{1}{{34}} + \dfrac{1}{{36}} + \dfrac{1}{{38}} + \dfrac{1}{{40}}} \right) = 0\\
\Rightarrow x = 424\\
\text{Vậy}\,x = 424\\
k)Dkxd:x \ne 8\\
\dfrac{3}{{2x - 16}} + \dfrac{{3x - 20}}{{x - 8}} + \dfrac{1}{8} = \dfrac{{13x - 102}}{{3x - 24}}\\
\Rightarrow \dfrac{{3.12 + 24.\left( {3x - 20} \right) + 3\left( {x - 8} \right)}}{{24\left( {x - 8} \right)}}\\
= \dfrac{{8\left( {13x - 102} \right)}}{{24\left( {x - 8} \right)}}\\
\Rightarrow 36 + 72x - 480 + 3x - 24 = 104x - 816\\
\Rightarrow 29x = 348\\
\Rightarrow x = 12\left( {tmdk} \right)\\
\text{Vậy}\,x = 12
\end{array}$
