Đáp án:
\[x = \frac{{ - 4 + \sqrt {21} }}{{10}}\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{1}{{\sqrt {2x + 1} - \sqrt {3x} }} = \frac{{\sqrt {3x + 2} }}{{1 - x}}\\
\Leftrightarrow \frac{{\sqrt {2x + 1} + \sqrt {3x} }}{{\left( {\sqrt {2x + 1} - \sqrt {3x} } \right)\left( {\sqrt {2x + 1} + \sqrt {3x} } \right)}} = \frac{{\sqrt {3x + 2} }}{{1 - x}}\\
\Leftrightarrow \frac{{\sqrt {2x + 1} + \sqrt {3x} }}{{\left( {2x + 1} \right) - 3x}} = \frac{{\sqrt {3x + 2} }}{{1 - x}}\\
\Leftrightarrow \frac{{\sqrt {2x + 1} + \sqrt {3x} }}{{1 - x}} = \frac{{\sqrt {3x + 2} }}{{1 - x}}\\
\Leftrightarrow \sqrt {2x + 1} + \sqrt {3x} = \sqrt {3x + 2} \\
\Leftrightarrow {\left( {\sqrt {2x + 1} + \sqrt {3x} } \right)^2} = 3x + 2\\
\Leftrightarrow 2x + 1 + 3x + 2\sqrt {\left( {2x + 1} \right).3x} = 3x + 2\\
\Leftrightarrow 2\sqrt {3x\left( {2x + 1} \right)} = 1 - 2x\,\,\,\,\,\left( {x \le \frac{1}{2}} \right)\\
\Leftrightarrow 4.3x.\left( {2x + 1} \right) = {\left( {1 - 2x} \right)^2}\\
\Leftrightarrow 24{x^2} + 12x = 4{x^2} - 4x + 1\\
\Leftrightarrow 20{x^2} + 16x - 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 4 + \sqrt {21} }}{{10}}\left( {t/m} \right)\\
x = \frac{{ - 4 - \sqrt {21} }}{{10}}\left( L \right)
\end{array} \right.
\end{array}\)
Vậy \(x = \frac{{ - 4 + \sqrt {21} }}{{10}}\)
