Đáp án:
$\begin{array}{l}
a){\left( {x - 2} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)\\
+ 6 + {\left( {x + 1} \right)^2} = 15\\
\Rightarrow {x^3} - 3.{x^2}.2 + 3.x.4 - 8 - \left( {{x^3} - {3^3}} \right)\\
+ 6 + {x^2} + 2x + 1 = 15\\
\Rightarrow {x^3} - 6{x^2} + 12x - 8 - {x^3}\\
+ 27 + {x^2} + 2x - 8 = 0\\
\Rightarrow - 5{x^2} + 14x + 9 = 0\\
\Rightarrow 5{x^2} - 14x - 9 = 0\\
\Rightarrow x = \frac{{12 \pm \sqrt {94} }}{5}\\
b)6{x^2} - 6x\left( { - 2 + x} \right) = 36\\
\Rightarrow 6{x^2} + 12x - 6{x^2} = 36\\
\Rightarrow 12x = 36\\
\Rightarrow x = 3\\
c){\left( {x + 2} \right)^2} + {\left( {x - 3} \right)^2} - 2\left( {x - 1} \right)\left( {x + 1} \right) = 9\\
\Rightarrow {x^2} + 4x + 4 + {x^2} - 6x + 9 - 2\left( {{x^2} - 1} \right) = 9\\
\Rightarrow 2{x^2} - 2x + 13 - 2{x^2} + 2 - 9 = 0\\
\Rightarrow - 2x + 6 = 0\\
\Rightarrow x = 3\\
d){\left( {x + 5} \right)^2} - 9 = 0\\
\Rightarrow {\left( {x + 5} \right)^2} = 9\\
\Rightarrow \left[ \begin{array}{l}
x + 5 = 3\\
x + 5 = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\\
x = - 8
\end{array} \right.\\
e){\left( {x - 2} \right)^3} = {x^3} + 6{x^2} + 7\\
\Rightarrow {x^3} - 6{x^2} + 12x - 8 = {x^3} + 6{x^2} + 7\\
\Rightarrow 12{x^2} - 12x + 15 = 0\\
\Rightarrow 4{x^2} - 4x + 5 = 0\\
\Rightarrow \left( {4{x^2} - 4x + 1} \right) + 4 = 0\\
\Rightarrow {\left( {2x - 1} \right)^2} + 4 = 0\left( {\text{vô}\,\text{nghiệm}} \right)
\end{array}$
Vậy pt vô nghiệm.
