Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 3 }}{2}\\
x = \dfrac{{ - 7 + 3\sqrt 5 }}{2}\\
x = \dfrac{{ - 7 - 3\sqrt 5 }}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
16{x^2} + 64x - 3\left| {16x + 2} \right| + 4 = 0\\
\to 16{x^2} + 64x + 4 = 3\left| {16x + 2} \right|\\
\to \left[ \begin{array}{l}
16{x^2} + 64x + 4 = 3\left( {16x + 2} \right)\left( {DK:x \ge \dfrac{1}{8}} \right)\\
16{x^2} + 64x + 4 = - 3\left( {16x + 2} \right)\left( {DK:x < \dfrac{1}{8}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
16{x^2} + 16x - 8 = 0\\
16{x^2} + 112x + 16 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 3 }}{2}\left( {TM} \right)\\
x = \dfrac{{ - 1 - \sqrt 3 }}{2}\left( l \right)\\
x = \dfrac{{ - 7 + 3\sqrt 5 }}{2}\left( {TM} \right)\\
x = \dfrac{{ - 7 - 3\sqrt 5 }}{2}\left( {TM} \right)
\end{array} \right.
\end{array}\)
