Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{ - 6 \pm 5\sqrt 2 }}{{14}}\\
x = \frac{{ - 8 \pm \sqrt {46} }}{{14}}
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge - \frac{9}{4}\)
Ta có:
\(\begin{array}{l}
7{x^2} + 7x = \sqrt {\frac{{4x + 9}}{{28}}} \\
\Leftrightarrow {\left( {7{x^2} + 7x} \right)^2} = \frac{{4x + 9}}{{28}}\\
\Leftrightarrow 49{x^4} + 98{x^3} + 49{x^2} - \frac{{4x + 9}}{{28}} = 0\\
\Leftrightarrow 1372{x^4} + 2744{x^3} + 1372{x^2} - 4x - 9 = 0\\
\Leftrightarrow \left( {1372{x^4} + 1176{x^3} - 98{x^2}} \right) + \left( {1568{x^3} + 1344{x^2} - 112x} \right) + \left( {126{x^2} + 108x - 9} \right) = 0\\
\Leftrightarrow 98{x^2}\left( {14{x^2} + 12x - 1} \right) + 112x\left( {14{x^2} + 12x - 1} \right) + 9\left( {14{x^2} + 12x - 1} \right) = 0\\
\Leftrightarrow \left( {14{x^2} + 12x - 1} \right)\left( {98{x^2} + 112x + 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
14{x^2} + 12x - 1 = 0\\
98{x^2} + 112x + 9 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 6 \pm 5\sqrt 2 }}{{14}}\left( {t/m} \right)\\
x = \frac{{ - 8 \pm \sqrt {46} }}{{14}}\left( {t/m} \right)
\end{array} \right.
\end{array}\)
