Đáp án:
$\begin{array}{l}
1)a)x = \pm \sqrt 5 \\
b)x = - 3 \pm \sqrt {19} \\
c)x = 3;x = - \dfrac{1}{3}\\
d)x = \dfrac{{145}}{3}\\
B2)a)Dkxd:x > 0;x\# 9\\
C = \left( {\dfrac{{x - 3\sqrt x }}{{x - 6\sqrt x + 9}} - \dfrac{{2\sqrt x - 1}}{{x - 3\sqrt x }}} \right).\dfrac{{x - 9}}{{\sqrt x + 3}}\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{{{\left( {\sqrt x - 3} \right)}^2}}} - \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right).\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}}\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right).\left( {\sqrt x - 3} \right)\\
= \dfrac{{\sqrt x .\sqrt x - 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 3} \right)}}.\left( {\sqrt x - 3} \right)\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
b)x = 0,25 = \dfrac{1}{4}\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = \dfrac{1}{2}\\
\Leftrightarrow C = \dfrac{{{{\left( {\dfrac{1}{2} - 1} \right)}^2}}}{{\dfrac{1}{2}}} = \dfrac{1}{2}\\
c)C = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} \ge 0\\
\Leftrightarrow GTNN:C = 0\,khi:x = 1
\end{array}$
