Đáp án:
$S = \left\{ {1;\dfrac{{1 + \sqrt {13} }}{2};\sqrt[3]{{\dfrac{{ - 31}}{9}}}} \right\}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
6x\sqrt {2{x^3} + 7} = 6{x^3} + 2x + 22 - 4\sqrt {2{x^3} + 7} \left( {DK:2{x^3} + 7 \ge 0} \right)\\
\Leftrightarrow \left( {6x + 4} \right)\sqrt {2{x^3} + 7} = 6{x^3} + 2x + 22\\
\Leftrightarrow \left( {6x + 3} \right)\sqrt {2{x^3} + 7} + \sqrt {2{x^3} + 7} = 6{x^3} + 2x + 22\\
\Leftrightarrow 3\left( {2x + 1} \right)\sqrt {2{x^3} + 7} + \sqrt {2{x^3} + 7} = 3\left( {2{x^3} + 7} \right) + 2x + 1\\
\Leftrightarrow 3\left( {2x + 1} \right)\sqrt {2{x^3} + 7} - 3\left( {2{x^3} + 7} \right) + \sqrt {2{x^3} + 7} - \left( {2x + 1} \right) = 0\\
\Leftrightarrow 3\sqrt {2{x^3} + 7} \left( {2x + 1 - \sqrt {2{x^3} + 7} } \right) + \sqrt {2{x^3} + 7} - \left( {2x + 1} \right) = 0\\
\Leftrightarrow \left( {2x + 1 - \sqrt {2{x^3} + 7} } \right)\left( {3\sqrt {2{x^3} + 7} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 - \sqrt {2{x^3} + 7} = 0\\
3\sqrt {2{x^3} + 7} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {2{x^3} + 7} = 2x + 1\\
\sqrt {2{x^3} + 7} = \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x + 1 \ge 0\\
2{x^3} + 7 = {\left( {2x + 1} \right)^2}
\end{array} \right.\\
2{x^3} + 7 = \dfrac{1}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge \dfrac{{ - 1}}{2}\\
2{x^3} - 4{x^2} - 4x + 6 = 0
\end{array} \right.\\
{x^3} = \dfrac{{ - 31}}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge \dfrac{{ - 1}}{2}\\
2\left( {x - 1} \right)\left( {{x^2} - x - 3} \right) = 0
\end{array} \right.\\
x = \sqrt[3]{{\dfrac{{ - 31}}{9}}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge \dfrac{{ - 1}}{2}\\
\left[ \begin{array}{l}
x = 1\left( c \right)\\
x = \dfrac{{1 + \sqrt {13} }}{2}\left( c \right)\\
x = \dfrac{{1 - \sqrt {13} }}{2}\left( l \right)
\end{array} \right.
\end{array} \right.\\
x = \sqrt[3]{{\dfrac{{ - 31}}{9}}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{1 + \sqrt {13} }}{2}\\
x = \sqrt[3]{{\dfrac{{ - 31}}{9}}}
\end{array} \right.
\end{array}$
Vậy tập nghiệm của phương trình là: $S = \left\{ {1;\dfrac{{1 + \sqrt {13} }}{2};\sqrt[3]{{\dfrac{{ - 31}}{9}}}} \right\}$
