Đáp án:
$\begin{array}{l}
C3)\\
a)\sqrt {{{\left( {x - 2} \right)}^2}} - 5 = 1\\
\Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = 6\\
\Leftrightarrow \left| {x - 2} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 6\\
x - 2 = - 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = - 4
\end{array} \right.\\
Vậy\,x = 8;x = - 4\\
b)Dkxd:x \ge 0\\
\sqrt {49x} - \sqrt {36x} = 5\\
\Leftrightarrow 7\sqrt x - 6\sqrt x = 5\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = 25\left( {tmdk} \right)\\
Vậy\,x = 25\\
C4)a)\sqrt {{x^2}} < 2\\
\Leftrightarrow \left| x \right| < 2\\
\Leftrightarrow - 2 < x < 2\\
Vậy\, - 2 < x < 2\\
b)x - \sqrt {5y} + \sqrt {5x} - y\\
= x - y + \sqrt {5x} - \sqrt {5y} \\
= \left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right) + \sqrt 5 \left( {\sqrt x - \sqrt y } \right)\\
= \left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y + \sqrt 5 } \right)\\
C5)a)\\
A = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{x - 2\sqrt x + 1}}\\
= \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)A - 1\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} - 1\\
= \dfrac{{\sqrt x - 1 - \sqrt x }}{{\sqrt x }}\\
= \dfrac{{ - 1}}{{\sqrt x }}\\
Do:\sqrt x > 0\\
\Leftrightarrow \dfrac{{ - 1}}{{\sqrt x }} < 0\\
\Leftrightarrow A < 1\\
Vậy\,A < 1
\end{array}$
