Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > - \frac{1}{3}\\
\frac{1}{{x\sqrt {3 + x} }} + \frac{{2\sqrt x }}{{\sqrt {3x + 1} }} = \frac{3}{2}\\
\to 2\sqrt {3x + 1} + 4x\sqrt {3x + {x^2}} = 3x\sqrt {9x + 3 + 3{x^2} + x} \\
\to 2\sqrt {3x + 1} + 4x\sqrt {3x + {x^2}} = 3x\sqrt {3{x^2} + 10x + 3} \\
\to 4\left( {3x + 1} \right) + 16{x^2}\left( {3x + {x^2}} \right) = 9{x^2}\left( {3{x^2} + 10x + 3} \right)\\
\to 12x + 4 + 48{x^3} + 16{x^4} = 27{x^4} + 90{x^3} + 27{x^2}\\
\to 11{x^4} + 42{x^3} + 27{x^2} - 12x - 4 = 0\\
\left[ \begin{array}{l}
x = 0,4415675333\left( {TM} \right)\\
x = - 0,2459473225\left( {TM} \right)\\
x = - 1,182690437\left( l \right)\\
x = 2,831111592\left( l \right)
\end{array} \right.
\end{array}\)