Đáp án:
$\begin{array}{l}
a)\sqrt x = \dfrac{9}{{16}}\\
\Leftrightarrow x = {\left( {\dfrac{9}{{16}}} \right)^2}\\
\Leftrightarrow x = \dfrac{{81}}{{256}}\\
Vậy\,x = \dfrac{{81}}{{256}}\\
b)\sqrt {\dfrac{9}{{16}}} = x\\
\Leftrightarrow \dfrac{3}{4} = x\\
Vậy\,x = \dfrac{3}{4}\\
c)\left( {{x^2} - 7} \right)\left( {{x^2} + 2} \right) = 0\\
\Leftrightarrow {x^2} - 7 = 0\left( {do:{x^2} + 2 > 0} \right)\\
\Leftrightarrow {x^2} = 7\\
\Leftrightarrow x = \sqrt 7 ;x = - \sqrt 7 \\
Vậy\,x = \sqrt 7 ;x = - \sqrt 7
\end{array}$
$\begin{array}{l}
d)Dkxd:x \ge 0\\
\sqrt {2,25} + \left( {\sqrt x - \dfrac{3}{4}} \right).\dfrac{4}{7} = \dfrac{{11}}{6}\\
\Leftrightarrow 1,5 + \dfrac{4}{7}\sqrt x - \dfrac{3}{7} = \dfrac{{11}}{6}\\
\Leftrightarrow \dfrac{4}{7}\sqrt x = \dfrac{{11}}{6} + \dfrac{3}{7} - 1,5\\
\Leftrightarrow \dfrac{4}{7}\sqrt x = \dfrac{{16}}{{21}}\\
\Leftrightarrow \sqrt x = \dfrac{4}{3}\\
\Leftrightarrow x = \dfrac{{16}}{9}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{16}}{9}\\
e)Dkxd:x > 0\\
\sqrt {289} - 5\left| {\dfrac{6}{{25}}:\sqrt x - \sqrt {49} } \right| = - 15\\
\Leftrightarrow 17 - 5.\left| {\dfrac{6}{{25}}:\sqrt x - 7} \right| = - 15\\
\Leftrightarrow 5\left| {\dfrac{6}{{25}}:\sqrt x - 7} \right| = 32\\
\Leftrightarrow \left| {\dfrac{6}{{25}}:\sqrt x - 7} \right| = \dfrac{{32}}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{6}{{25}}:\sqrt x - 7 = \dfrac{{32}}{5}\\
\dfrac{6}{{25}}:\sqrt x - 7 = - \dfrac{{32}}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{6}{{335}}\\
\sqrt x = \dfrac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{36}}{{112225}}\\
x = \dfrac{4}{{25}}
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{36}}{{112225}};x = \dfrac{4}{{25}}
\end{array}$
