Giải thích các bước giải:
ĐKXĐ: \(x \ge - 3\)
Ta có:
\(\begin{array}{l}
2\sqrt {x + 3} = 9{x^2} - x - 4\\
\Leftrightarrow 4\left( {x + 3} \right) = {\left( {9{x^2} - x - 4} \right)^2}\\
\Leftrightarrow 4x + 12 = 81{x^4} + {x^2} + 16 - 18{x^3} + 8x - 72{x^2}\\
\Leftrightarrow 81{x^4} - 18{x^3} - 71{x^2} + 4x + 4 = 0\\
\Leftrightarrow \left( {81{x^4} - 81{x^3}} \right) + \left( {63{x^3} - 63{x^2}} \right) - \left( {8{x^2} - 8x} \right) - \left( {4x - 4} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {81{x^3} + 63{x^2} - 8x - 4} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left[ {\left( {81{x^3} + 18{x^2}} \right) + \left( {45{x^2} + 10x} \right) - \left( {18x + 4} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {9x + 2} \right)\left( {9{x^2} + 5x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \frac{2}{9}\\
x = \frac{{ - 5 \pm \sqrt {97} }}{{18}}
\end{array} \right.
\end{array}\)
