Đáp án:
$\begin{array}{l}
b)\left( {5 + \sqrt 2 } \right){x^2} + \left( {5 - \sqrt 2 } \right).x - 10 = 0\\
\Rightarrow \left( {5 + \sqrt 2 } \right){x^2} - \left( {5 + \sqrt 2 } \right)x + 10x - 10 = 0\\
\Rightarrow \left( {x - 1} \right)\left( {\left( {5 + \sqrt 2 } \right)x + 10} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{ - 10}}{{5 + \sqrt 2 }} = \frac{{ - 10\left( {5 - \sqrt 2 } \right)}}{{25 - 2}} = \frac{{10\sqrt 2 - 50}}{{23}}
\end{array} \right.\\
d)36{x^4} - 13{x^2} + 1 = 0\\
\Rightarrow 36{x^4} - 4{x^2} - 9{x^2} + 1 = 0\\
\Rightarrow \left( {9{x^2} - 1} \right)\left( {4{x^2} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = \frac{1}{9}\\
{x^2} = \frac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \pm \frac{1}{3}\\
x = \pm \frac{1}{2}
\end{array} \right.\\
g)5x + 2\sqrt x - 16 = 0\\
\Rightarrow 5x + 10\sqrt x - 8\sqrt x - 16 = 0\\
\Rightarrow \left( {\sqrt x + 2} \right)\left( {5\sqrt x - 8} \right) = 0\\
\Rightarrow \sqrt x = \frac{8}{5}\left( {do:\sqrt x + 2 > 0} \right)\\
\Rightarrow x = \frac{{64}}{{25}}
\end{array}$
