Đáp án:
c) \(\left[ \begin{array}{l}
x = 5\\
x = - 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a){\left( {2{x^2} - 3} \right)^2} = 4{\left( {x - 1} \right)^2}\\
\to \left| {2{x^2} - 3} \right| = 2\left| {x - 1} \right|\\
\to \left[ \begin{array}{l}
2x - 2 = 2{x^2} - 3\left( {DK:x > 1} \right)\\
2x - 2 = - 2{x^2} + 3\left( {DK:x \le 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2{x^2} - 2x - 1 = 0\\
2{x^2} + 3x - 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt 3 }}{2}\\
x = \dfrac{{1 - \sqrt 3 }}{2}\left( l \right)\\
x = 1\\
x = - \dfrac{5}{2}
\end{array} \right.\\
b)2x{\left( {3x - 1} \right)^2} - \left( {9{x^2} - 1} \right) = 0\\
\to 2x{\left( {3x - 1} \right)^2} - \left( {3x - 1} \right)\left( {3x + 1} \right) = 0\\
\to \left( {3x - 1} \right)\left[ {2x\left( {3x - 1} \right) - 3x - 1} \right] = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
6{x^2} - 2x - 3x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = 1\\
x = - \dfrac{1}{6}
\end{array} \right.\\
c)3{x^2} - 14\left| x \right| - 5 = 0\\
\to \left[ \begin{array}{l}
\left| x \right| = 5\\
\left| x \right| = - \dfrac{1}{3}\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 5
\end{array} \right.
\end{array}\)
