Đáp án:
$\begin{array}{l}
6)Đặt:{x^2} = t\left( {t \ge 0} \right)\\
\Rightarrow 5{t^2} - 3t + \dfrac{7}{{16}} = 0\\
\left( {a = 5;b = - 3;c = \dfrac{7}{{16}}} \right)\\
\Rightarrow \Delta = {\left( { - 3} \right)^2} - 4.5.\dfrac{7}{{16}} = \dfrac{1}{4}\\
\Rightarrow \left\{ \begin{array}{l}
{t_1} = \dfrac{{3 + \sqrt {\dfrac{1}{4}} }}{{2.5}} = \dfrac{7}{{20}}\\
{t_2} = \dfrac{{3 - \sqrt {\dfrac{1}{4}} }}{{2.5}} = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{ \pm \sqrt {35} }}{{10}}\\
x = \pm \dfrac{1}{2}
\end{array} \right.\\
7) - {x^4} + 6{x^2} = 0\\
\Rightarrow {x^2}\left( { - {x^2} + 6} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 6 \\
x = - \sqrt 6
\end{array} \right.\\
8) - {x^4} + 5{x^2} = 0\\
\Rightarrow {x^2}\left( { - {x^2} + 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 5 \\
x = - \sqrt 5
\end{array} \right.\\
9) - {x^4} + 81 = 0\\
\Rightarrow {x^4} = 81\\
\Rightarrow {\left( {{x^2}} \right)^2} = 81\\
\Rightarrow {x^2} = 9\left( {do:{x^2} \ge 0} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x = - 3
\end{array} \right.\\
10) - {x^4} + 25 = 0\\
\Rightarrow {x^4} = 25\\
\Rightarrow {\left( {{x^2}} \right)^2} = 25\\
\Rightarrow {x^2} = 5\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 5 \\
x = - \sqrt 5
\end{array} \right.
\end{array}$
