Đáp án:
\[x = 2\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
{x^2} + 28x + 4 \ge 0\\
x > 1
\end{array} \right. \Leftrightarrow x > 1\)
Ta có:
\(\begin{array}{l}
\frac{{\sqrt {{x^2} + 28x + 4} }}{{x + 2}} + 8 = \frac{{x + 4}}{{\sqrt {x - 1} }} + 2x\\
\Leftrightarrow \frac{{\sqrt {{x^2} + 28x + 4} }}{{x + 2}} - 2 = \left( {\frac{{x + 4}}{{\sqrt {x - 1} }} - 6} \right) + 2x - 4\\
\Leftrightarrow \frac{{\sqrt {{x^2} + 28x + 4} - 2\left( {x + 2} \right)}}{{x + 2}} = \frac{{\left( {x + 4} \right) - 6\sqrt {x - 1} }}{{\sqrt {x - 1} }} + 2\left( {x - 2} \right)\\
\Leftrightarrow \frac{{{x^2} + 28x + 4 - 4\left( {{x^2} + 4x + 4} \right)}}{{x + 2}} = \frac{{{x^2} + 8x + 16 - 36\left( {x - 1} \right)}}{{\sqrt {x - 1} \left( {x + 4 + 6\sqrt {x - 1} } \right)}} + 2\left( {x - 2} \right)\\
\Leftrightarrow \frac{{ - 3{x^2} + 12x - 12}}{{x + 2}} = \frac{{{x^2} - 28x + 52}}{{\sqrt {x - 1} \left( {x + 4 + 6\sqrt {x - 1} } \right)}} + 2\left( {x - 2} \right)\\
\Leftrightarrow \frac{{ - 3{{\left( {x - 2} \right)}^2}}}{{x + 2}} = \frac{{\left( {x - 2} \right)\left( {x - 26} \right)}}{{\sqrt {x - 1} \left( {x + 4 + 6\sqrt {x - 1} } \right)}} + 2\left( {x - 2} \right)\\
\Leftrightarrow x = 2
\end{array}\)
