Đáp án:
$\begin{array}{l}
c)\dfrac{{\sqrt {3 - 2\sqrt 2 } }}{{\sqrt {17 - 12\sqrt 2 } }} - \dfrac{{\sqrt {3 + 2\sqrt 2 } }}{{\sqrt {17 + 12\sqrt 2 } }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} }}{{\sqrt {9 - 2.3.2\sqrt 2 + 8} }} - \dfrac{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}{{\sqrt {9 + 2.3.2\sqrt 2 + 8} }}\\
= \dfrac{{\sqrt 2 - 1}}{{\sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} }} - \dfrac{{\sqrt 2 + 1}}{{\sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} }}\\
= \dfrac{{\sqrt 2 - 1}}{{3 - 2\sqrt 2 }} - \dfrac{{\sqrt 2 + 1}}{{3 + 2\sqrt 2 }}\\
= \dfrac{1}{{\sqrt 2 - 1}} - \dfrac{1}{{\sqrt 2 + 1}}\\
= \dfrac{{\sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\\
= \dfrac{2}{{2 - 1}}\\
= 2\\
B3)a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x\# 4\\
x\# 9
\end{array} \right.\\
A = \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{3\sqrt x - 3}}{{x - 5\sqrt x + 6}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right) - 3\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 4 - \left( {x - 2\sqrt x - 3} \right) - 3\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 1}}{{\sqrt x - 3}}\\
= \dfrac{1}{{3 - \sqrt x }}\\
b)3A = \dfrac{3}{{3 - \sqrt x }} \in Z\\
\Leftrightarrow 3 - \sqrt x \in \left\{ { - 3; - 1;1;3} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {6;4;2;0} \right\}\\
\Leftrightarrow x \in \left\{ {0;4;16;36} \right\}\\
Do:x\# 4;x\# 9\\
\Leftrightarrow x \in \left\{ {0;16;36} \right\}\\
Vay\,x \in \left\{ {0;16;36} \right\}
\end{array}$
