Đáp án:
\[x = 6\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge \frac{2}{3}\)
Ta có:
\(\begin{array}{l}
9.\left( {\sqrt {4x + 1} - \sqrt {3x - 2} } \right) = x + 3\\
\Leftrightarrow 9.\frac{{\left( {4x + 1} \right) - \left( {3x - 2} \right)}}{{\sqrt {4x + 1} + \sqrt {3x - 2} }} = x + 3\\
\Leftrightarrow 9.\frac{{x + 3}}{{\sqrt {4x + 1} + \sqrt {3x - 1} }} = x + 3\\
x \ge \frac{2}{3} \Rightarrow x + 3 > 0\\
\Rightarrow \frac{9}{{\sqrt {4x + 1} + \sqrt {3x - 2} }} = 1\\
\Leftrightarrow \sqrt {4x + 1} + \sqrt {3x - 2} = 9\\
\Leftrightarrow \left( {\sqrt {4x + 1} - 5} \right) + \left( {\sqrt {3x - 2} - 4} \right) = 0\\
\Leftrightarrow \frac{{4x + 1 - 25}}{{\sqrt {4x + 1} + 5}} + \frac{{3x - 2 - 16}}{{\sqrt {3x - 2} + 4}} = 0\\
\Leftrightarrow \left( {x - 6} \right)\left( {\frac{4}{{\sqrt {4x + 1} + 5}} + \frac{3}{{\sqrt {3x - 2} + 4}}} \right) = 0\\
\Leftrightarrow x - 6 = 0\\
\Leftrightarrow x = 6
\end{array}\)
