Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
\sqrt a + 3 \ne 0\\
a + \sqrt a - 6 \ne 0\\
2 - \sqrt a \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
a \ne 4
\end{array} \right.\\
H = \dfrac{{\sqrt a + 2}}{{\sqrt a + 3}} - \dfrac{5}{{a + \sqrt a - 6}} + \dfrac{1}{{2 - \sqrt a }}\\
= \dfrac{{\sqrt a + 2}}{{\sqrt a + 3}} - \dfrac{5}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}} - \dfrac{1}{{\sqrt a - 2}}\\
= \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right) - 5 - \left( {\sqrt a + 3} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \dfrac{{\left( {a - 4} \right) - 5 - \left( {\sqrt a + 3} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \dfrac{{a - \sqrt a - 12}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \dfrac{{\left( {\sqrt a - 4} \right)\left( {\sqrt a + 3} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \dfrac{{\sqrt a - 4}}{{\sqrt a - 2}}\\
b,\\
H < 2\\
\Leftrightarrow \dfrac{{\sqrt a - 4}}{{\sqrt a - 2}} < 2\\
\Leftrightarrow \dfrac{{\sqrt a - 4}}{{\sqrt a - 2}} - 2 < 0\\
\Leftrightarrow \dfrac{{\left( {\sqrt a - 4} \right) - 2.\left( {\sqrt a - 2} \right)}}{{\sqrt a - 2}} < 0\\
\Leftrightarrow \dfrac{{ - \sqrt a }}{{\sqrt a - 2}} < 0\\
\Leftrightarrow \dfrac{{\sqrt a }}{{\sqrt a - 2}} > 0\\
\Leftrightarrow \sqrt a - 2 > 0\\
\Leftrightarrow a > 4\\
c,\\
x = 7 + 4\sqrt 3 = 4 + 2.2.\sqrt 3 + 3 = {\left( {2 + \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt x = 2 + \sqrt 3 \\
\Rightarrow H = \dfrac{{\left( {2 + \sqrt 3 } \right) - 4}}{{\left( {2 + \sqrt 3 } \right) - 2}} = \dfrac{{\sqrt 3 - 2}}{{\sqrt 3 }} = \dfrac{{3 - 2\sqrt 3 }}{3}\\
d,\\
H = 5\\
\Leftrightarrow \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}} = 5\\
\Leftrightarrow \sqrt x - 4 = 5.\left( {\sqrt x - 2} \right)\\
\Leftrightarrow \sqrt x - 4 = 5\sqrt x - 10\\
\Leftrightarrow 4\sqrt x = 6\\
\Leftrightarrow \sqrt x = \dfrac{3}{2}\\
\Leftrightarrow x = \dfrac{9}{4}
\end{array}\)
