Đáp án:
$\begin{array}{l}
1)25{x^2} - 9 = 0\\
\Leftrightarrow \left( {5x - 3} \right)\left( {5x + 3} \right) = 0\\
\Leftrightarrow x = \dfrac{3}{5};x = - \dfrac{3}{5}\\
Vậy\,x = \dfrac{3}{5};x = - \dfrac{3}{5}\\
2){\left( {5x - 2} \right)^2} - 64{x^2} = 0\\
\Leftrightarrow \left( {5x - 2 - 8x} \right)\left( {5x - 2 + 8x} \right) = 0\\
\Leftrightarrow \left( { - 3x - 2} \right)\left( {13x - 2} \right) = 0\\
\Leftrightarrow x = - \dfrac{2}{3};x = \dfrac{3}{{13}}\\
Vậy\,x = - \dfrac{2}{3};x = \dfrac{3}{{13}}\\
4){x^2} - 4x + 4 = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
5){\left( {5 - 2x} \right)^2} - 16 = 0\\
\Leftrightarrow \left( {5 - 2x - 4} \right)\left( {5 - 2x + 4} \right) = 0\\
\Leftrightarrow \left( {1 - 2x} \right)\left( {9 - 2x} \right) = 0\\
\Leftrightarrow x = \dfrac{1}{2};x = \dfrac{9}{2}\\
Vậy\,x = \dfrac{1}{2};x = \dfrac{9}{2}\\
6){\left( {2x + 1} \right)^3} = {\left( {x - 5} \right)^3}\\
\Leftrightarrow 2x + 1 = x - 5\\
\Leftrightarrow x = - 6\\
Vậy\,x = - 6\\
7){\left( {3 - 4x} \right)^3} = - {\left( {3x + 2} \right)^3}\\
\Leftrightarrow {\left( {3 - 4x} \right)^3} = {\left( { - 3x - 2} \right)^3}\\
\Leftrightarrow 3 - 4x = - 3x - 2\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
8){\left( {x + 1} \right)^3} - {\left( {5 - x} \right)^3} = 0\\
\Leftrightarrow {\left( {x + 1} \right)^3} = {\left( {5 - x} \right)^3}\\
\Leftrightarrow x + 1 = 5 - x\\
\Leftrightarrow 2x = 4\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
9){\left( {5 + \dfrac{1}{2}x} \right)^3} + {\left( {x - 2} \right)^3} = 0\\
\Leftrightarrow {\left( {5 + \dfrac{1}{2}x} \right)^3} = {\left( {2 - x} \right)^3}\\
\Leftrightarrow 5 + \dfrac{1}{2}x = 2 - x\\
\Leftrightarrow \dfrac{3}{2}x = - 3\\
\Leftrightarrow x = - 2\\
Vậy\,x = - 2
\end{array}$