Đáp án:
$\begin{array}{l}
1)a){\left( {2x + 1} \right)^2} - 4{\left( {x + 2} \right)^2} = 9\\
\Leftrightarrow 4{x^2} + 4x + 1 - 4\left( {{x^2} + 4x + 4} \right) = 9\\
\Leftrightarrow 4{x^2} + 4x + 1 - 4{x^2} - 16x - 16 = 9\\
\Leftrightarrow - 12x = 24\\
\Leftrightarrow x = - 2\\
Vậy\,x = - 2\\
b){\left( {x + 3} \right)^2} - \left( {x - 4} \right)\left( {x + 8} \right) = 1\\
\Leftrightarrow {x^2} + 6x + 9 - \left( {{x^2} + 4x - 32} \right) = 1\\
\Leftrightarrow {x^2} + 6x + 9 - {x^2} - 4x + 32 = 1\\
\Leftrightarrow 2x = - 40\\
\Leftrightarrow x = - 20\\
Vậy\,x = - 20\\
c)3{\left( {x + 2} \right)^2} + {\left( {2x - 1} \right)^2} - 7\left( {x + 3} \right)\left( {x - 3} \right) = 36\\
\Leftrightarrow 3\left( {{x^2} + 4x + 4} \right) + 4{x^2} - 4x + 1\\
- 7\left( {{x^2} - 9} \right) = 36\\
\Leftrightarrow 7{x^2} + 8x + 13 - 7{x^2} + 63 = 36\\
\Leftrightarrow 8x = - 40\\
\Leftrightarrow x = - 5\\
Vậy\,x = - 5\\
B2){x^2} + 2x + {y^2} - 6y + 4{z^2} - 4z + 11 = 0\\
\Leftrightarrow {x^2} + 2x + 1 + {y^2} - 6y + 9\\
+ 4{z^2} - 4z + 1 = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {2z - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 0\\
y - 3 = 0\\
2z - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 3\\
z = \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = - 1;y = 3;z = \dfrac{1}{2}
\end{array}$
