Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {{x^2} - 4x + 4} \right) - 16 = 0\\
\to {\left( {x - 2} \right)^2} = 16\\
\to \left[ \begin{array}{l}
x - 2 = 4\\
x - 2 = - 4
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.\\
b.\left( {4{x^2} - 4x + 1} \right) - 4 = 0\\
\to {\left( {2x - 1} \right)^2} = 4\\
\to \left[ \begin{array}{l}
2x - 1 = 2\\
2x - 1 = - 2
\end{array} \right. \to \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = - \frac{1}{2}
\end{array} \right.\\
c.{\left( {x + 1} \right)^2} = 4{\left( {x - 1} \right)^2}\\
\to \left[ \begin{array}{l}
x + 1 = 4\left( {x + 1} \right)\\
x + 1 = - 4\left( {x + 1} \right)
\end{array} \right. \to \left[ \begin{array}{l}
3x = - 3\\
5x = - 5
\end{array} \right. \to x = - 1\\
d.2{x^2} - 3\left( {4{x^2} - 12x + 9} \right) = 0\\
\to - 10{x^2} + 36x - 27 = 0\\
\to \left[ \begin{array}{l}
x = \frac{{18 + 3\sqrt 6 }}{{10}}\\
x = \frac{{18 - 3\sqrt 6 }}{{10}}
\end{array} \right.\\
e.{\left( {3x} \right)^2} = 4\\
\to \left[ \begin{array}{l}
3x = 2\\
3x = - 2
\end{array} \right. \to \left[ \begin{array}{l}
x = \frac{2}{3}\\
x = - \frac{2}{3}
\end{array} \right.\\
f.x\left( {\frac{{21}}{5}x + \frac{{273}}{{50}}} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \frac{{13}}{{10}}
\end{array} \right.
\end{array}\)