Đáp án:
$\begin{array}{l}
7)3x - 10\sqrt x + 3 = 0\left( {dkxd:x \ge 0} \right)\\
\Rightarrow 3x - 9\sqrt x - \sqrt x + 3 = 0\\
\Rightarrow \left( {\sqrt x - 3} \right)\left( {3\sqrt x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = \dfrac{1}{3}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 9\\
x = \dfrac{1}{9}
\end{array} \right.\left( {tmdk} \right)\\
\text{Vậy}\,x = 9\,\text{hoặc}\,x = \dfrac{1}{9}\\
8)x + 2\sqrt x - 3 = 0\\
\Rightarrow x - \sqrt x + 3\sqrt x - 3 = 0\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right) = 0\\
\Rightarrow \sqrt x = 1\\
\Rightarrow x = 1\left( {tmdk} \right)\\
\text{Vậy}\,x = 1\\
9)\sqrt {x + 6\sqrt x + 9} = x + 1\left( {dkxd:x \ge 0} \right)\\
\Rightarrow \sqrt {{{\left( {\sqrt x + 3} \right)}^2}} = x + 1\\
\Rightarrow \sqrt x + 3 = x + 1\\
\Rightarrow x - \sqrt x - 2 = 0\\
\Rightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) = 0\\
\Rightarrow \sqrt x - 2 = 0\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)\\
\text{Vậy}\,x = 4\\
10)Dkxd:x \ge \dfrac{5}{2}\\
\sqrt {1 + x} = \sqrt {2x - 5} \\
\Rightarrow 1 + x = 2x - 5\\
\Rightarrow x = 6\left( {tmdk} \right)\\
\text{Vậy}\,x = 6
\end{array}$
