Đáp án:
\(\left[ \begin{array}{l}
x = 16\\
x = \dfrac{{11 + 4\sqrt 7 }}{9}\\
x = 0,04914280529
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne \left\{ {1;4;9} \right\}\\
\left| M \right| = 3\sqrt x \\
\to \left[ \begin{array}{l}
M = 3\sqrt x \\
M = - 3\sqrt x
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{x + 3\sqrt x - 4}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} = 3\sqrt x \\
\dfrac{{x + 3\sqrt x - 4}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} = - 3\sqrt x
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{x + 3\sqrt x - 4 - 3\sqrt x \left( {x - 5\sqrt x + 6} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} = 0\\
\dfrac{{x + 3\sqrt x - 4 + 3\sqrt x \left( {x - 5\sqrt x + 6} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + 3\sqrt x - 4 - 3x\sqrt x + 15x - 18\sqrt x = 0\\
x + 3\sqrt x - 4 + 3x\sqrt x - 15x + 18\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3x\sqrt x + 16x - 15\sqrt x - 4 = 0\\
3x\sqrt x - 14x + 21\sqrt x - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {4 - \sqrt x } \right)\left( {3x - 4\sqrt x - 1} \right) = 0\\
\sqrt x = 0,2216817658
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 4\\
3x - 4\sqrt x - 1 = 0\\
\sqrt x = 0,2216817658
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 16\\
\sqrt x = \dfrac{{2 + \sqrt 7 }}{3}\\
\sqrt x = \dfrac{{2 - \sqrt 7 }}{3}\left( l \right)\\
x = {\left( {0,2216817658} \right)^2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 16\\
x = {\left( {\dfrac{{2 + \sqrt 7 }}{3}} \right)^2} = \dfrac{{11 + 4\sqrt 7 }}{9}\\
x = 0,04914280529
\end{array} \right.
\end{array}\)
