Đáp án:
$\begin{array}{l}
a){\left( {x + 2} \right)^2} - 9 = 0\\
\Leftrightarrow \left( {x + 2 - 3} \right)\left( {x + 2 + 3} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow x = 1;x = - 5\\
Vậy\,x = 1;x = - 5\\
b){x^2} - 2x + 1 = 25\\
\Leftrightarrow {\left( {x - 1} \right)^2} = {5^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 5\\
x - 1 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 4
\end{array} \right.\\
Vậy\,x = 6;x = - 4\\
c)4x\left( {x - 1} \right) - \left( {2x + 5} \right)\left( {2x - 5} \right) = 1\\
\Leftrightarrow 4{x^2} - 4x - 4{x^2} + 25 = 1\\
\Leftrightarrow 4x = 24\\
\Leftrightarrow x = 6\\
Vậy\,x = 6\\
d)\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - x\left( {{x^2} + 3} \right) = 17\\
\Leftrightarrow {x^3} + 27 - {x^3} - 3x = 17\\
\Leftrightarrow 3x = 10\\
\Leftrightarrow x = \frac{{10}}{3}\\
Vậy\,x = \frac{{10}}{3}\\
e)3{\left( {x + 2} \right)^2} + {\left( {2x - 1} \right)^2} - 7\left( {x + 3} \right)\left( {x - 3} \right) = 28\\
\Leftrightarrow 3\left( {{x^2} + 4x + 4} \right) + 4{x^2} - 4x + 1\\
- 7\left( {{x^2} - 9} \right) = 28\\
\Leftrightarrow 3{x^2} + 12x + 12 + 4{x^2} - 4x + 1\\
- 7{x^2} + 63 = 28\\
\Leftrightarrow 8x = - 48\\
\Leftrightarrow x = - 6\\
Vậy\,x = - 6
\end{array}$
