Đáp án:
$\begin{array}{l}
B7)\\
a)DKxd:x \ge 0;x\# 1\\
G = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{\sqrt x }}{{1 - \sqrt x }} + \dfrac{2}{{x - 1}}} \right)\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1 + \sqrt x \left( {\sqrt x + 1} \right) + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x + 2\sqrt x + 1}}\\
= \dfrac{{4\sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
b)G = \dfrac{8}{9}\\
\Leftrightarrow \dfrac{{4\sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}}} = \dfrac{8}{9}\\
\Leftrightarrow 9\sqrt x = 2{\left( {\sqrt x + 1} \right)^2}\\
\Leftrightarrow 9\sqrt x = 2x + 4\sqrt x + 2\\
\Leftrightarrow 2x - 5\sqrt x + 2 = 0\\
\Leftrightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\left( {tm} \right)\\
x = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{1}{4};x = 4\\
c)x = 17 + 12\sqrt 2 \\
= 9 + 2.3.2\sqrt 2 + 8\\
= {\left( {3 + 2\sqrt 2 } \right)^2}\\
\Leftrightarrow \sqrt x = 3 + 2\sqrt 2 \\
\Leftrightarrow G = \dfrac{{4\sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= \dfrac{{4.\left( {3 + 2\sqrt 2 } \right)}}{{{{\left( {3 + 2\sqrt 2 + 1} \right)}^2}}} = \dfrac{1}{2}\\
B8)\\
a)Dkxd:x \ge 0;x\# 9;x\# 4\\
H = \left( {\dfrac{{x - 3\sqrt x }}{{x - 9}} - 1} \right):\left( {\dfrac{{9 - x}}{{x + \sqrt x - 6}} - \dfrac{{\sqrt x - 3}}{{2 - \sqrt x }} - \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} - 1} \right)\\
:\left( {\dfrac{{9 - x + \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}} \right)\\
= \dfrac{{\sqrt x - \sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{9 - x + x - 9 - x + 4\sqrt x - 4}}\\
= \dfrac{{ - 3}}{1}.\dfrac{{\sqrt x - 2}}{{ - x + 4\sqrt x - 4}}\\
= 3.\dfrac{{\sqrt x - 2}}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\
= \dfrac{3}{{\sqrt x - 2}}\\
b)H < 0\\
\Leftrightarrow \dfrac{3}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \sqrt x - 2 < 0\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow x < 4\\
Vậy\,0 \le x < 4\\
c)H = \dfrac{3}{{\sqrt x - 2}}\\
Do:\sqrt x - 2 \ge - 2\\
\Leftrightarrow \dfrac{3}{{\sqrt x - 2}} \le \dfrac{{ - 3}}{2}\\
\Leftrightarrow H \le \dfrac{{ - 3}}{2}\\
\Leftrightarrow GTLN:H = - \dfrac{3}{2}\,khi:x = 0
\end{array}$
