Đáp án:
$\begin{array}{l}
19){\left( {2x - 1} \right)^2} - {\left( {5x - 5} \right)^2} = 0\\
\Leftrightarrow \left( {2x - 1 - 5x + 5} \right)\left( {2x - 1 + 5x - 5} \right) = 0\\
\Leftrightarrow \left( {4 - 3x} \right)\left( {7x - 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = \dfrac{6}{7}
\end{array} \right.\\
\Leftrightarrow C\\
20){\left( {2x + 1} \right)^2} - 4{\left( {x + 3} \right)^2} = 0\\
\Leftrightarrow {\left( {2x + 1} \right)^2} - {\left( {2x + 6} \right)^2} = 0\\
\Leftrightarrow \left( {2x + 1 - 2x - 6} \right)\left( {2x + 1 + 2x + 6} \right) = 0\\
\Leftrightarrow - 5.\left( {4x + 7} \right) = 0\\
\Leftrightarrow x = \dfrac{{ - 7}}{4}\\
\Leftrightarrow B\\
21)\\
\left( {x - 6} \right)\left( {x + 6} \right) - {\left( {x + 3} \right)^2} = 9\\
\Leftrightarrow {x^2} - 36 - {x^2} - 6x - 9 = 9\\
\Leftrightarrow 6x = - 36\\
\Leftrightarrow x = - 6\\
\Leftrightarrow D\\
22){\left( {3x - 1} \right)^2} + 2{\left( {x + 3} \right)^2} + 11\left( {1 + x} \right)\left( {1 - x} \right) = 6\\
\Leftrightarrow 9{x^2} - 6x + 1 + 2\left( {{x^2} + 6x + 9} \right)\\
+ 11\left( {1 - {x^2}} \right) = 6\\
\Leftrightarrow 11{x^2} + 6x + 19 + 11 - 11{x^2} = 6\\
\Leftrightarrow 6x = - 24\\
\Leftrightarrow x = - 4\\
\Leftrightarrow B
\end{array}$
