Đáp án:
$\begin{array}{l}
1){x^4} + 4{x^3} - 16x - 16 = 0\\
\Leftrightarrow {x^4} + 4{x^3} + 4{x^2} - 4{x^2} - 16x - 16 = 0\\
\Leftrightarrow {\left( {{x^2} + 2x} \right)^2} - {\left( {2x - 4} \right)^2} = 0\\
\Leftrightarrow \left( {{x^2} + 2x - 2x + 4} \right)\left( {{x^2} + 2x + 2x - 4} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 4} \right)\left( {{x^2} + 4x - 4} \right) = 0\\
\Leftrightarrow {x^2} + 4x - 4 = 0\\
\Leftrightarrow {x^2} + 4x + 4 = 8\\
\Leftrightarrow {\left( {x + 2} \right)^2} = 8\\
\Leftrightarrow x = - 2 \pm 2\sqrt 2 \\
Vậy\,x = - 2 \pm 2\sqrt 2 \\
2){\left( {2x - 3} \right)^2} - {\left( {x + 5} \right)^2} = 0\\
\Leftrightarrow \left( {2x - 3 - x - 5} \right)\left( {2x - 3 + x + 5} \right) = 0\\
\Leftrightarrow \left( {x - 8} \right)\left( {3x + 2} \right) = 0\\
\Leftrightarrow x = 8;x = - \dfrac{2}{3}\\
Vậy\,x = 8;x = - \dfrac{2}{3}\\
3){x^2} - 2x + {y^2} + 4y + 5 + {\left( {2z - 3} \right)^2} = 0\\
\Leftrightarrow {x^2} - 2x + 1 + {y^2} + 4y + 4\\
+ {\left( {2z - 3} \right)^2} = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {2z - 3} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 = 0\\
y + 2 = 0\\
2z - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = - 2\\
z = \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x = 1;y = - 2;z = \dfrac{3}{2}
\end{array}$
