Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\sqrt {3 + 2\sqrt 2 } - \sqrt {3 - 2\sqrt 2 } \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
= \left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)\\
= 2\\
2,\\
a - 5a\sqrt b + 6{a^2}\\
= a.\left( {1 - 5\sqrt b + 6a} \right)\\
3,\\
P = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right)\\
= \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{\left( {2\sqrt x - 9} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {2\sqrt x - 9} \right) - \left( {x - 9} \right) + \left( {2x - 3\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b,\\
P < 1 \Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} < 1\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} - 1 < 0\\
\Leftrightarrow \dfrac{{\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}} < 0\\
\Leftrightarrow \dfrac{4}{{\sqrt x - 3}} < 0\\
\Leftrightarrow \sqrt x < 3\\
\Leftrightarrow x < 9\\
\Rightarrow \left\{ \begin{array}{l}
0 \le x < 9\\
x \ne 4
\end{array} \right.
\end{array}\)
