Đáp án:
\(\begin{array}{l}
a,\\
P = \dfrac{{\sqrt a - 2}}{{\sqrt a - 4}}\\
b,\\
P = \dfrac{{6 - \sqrt 3 }}{{11}}\\
c,\\
a \in \left\{ {4;25;36} \right\}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
a - 7\sqrt a + 12 \ne 0\\
\sqrt a - 4 \ne 0\\
3 - \sqrt a \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
a \ne 16\\
a \ne 9
\end{array} \right.\\
a,\\
P = \dfrac{{2\sqrt a + 1}}{{a - 7\sqrt a + 12}} - \dfrac{{\sqrt a + 3}}{{\sqrt a - 4}} - \dfrac{{2\sqrt a + 1}}{{3 - \sqrt a }}\\
= \dfrac{{2\sqrt a + 1}}{{\left( {a - 3\sqrt a } \right) + \left( { - 4\sqrt a + 12} \right)}} - \dfrac{{\sqrt a + 3}}{{\sqrt a - 4}} + \dfrac{{2\sqrt a + 1}}{{\sqrt a - 3}}\\
= \dfrac{{2\sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 3} \right) - 4\left( {\sqrt a - 3} \right)}} - \dfrac{{\sqrt a + 3}}{{\sqrt a - 4}} + \dfrac{{2\sqrt a + 1}}{{\sqrt a - 3}}\\
= \dfrac{{2\sqrt a + 1}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}} - \dfrac{{\sqrt a + 3}}{{\sqrt a - 4}} + \dfrac{{2\sqrt a + 1}}{{\sqrt a - 3}}\\
= \dfrac{{\left( {2\sqrt a + 1} \right) - \left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right) + \left( {2\sqrt a + 1} \right)\left( {\sqrt a - 4} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{2\sqrt a + 1 - \left( {a - 9} \right) + \left( {2a - 7\sqrt a - 4} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{2\sqrt a + 1 - a + 9 + 2a - 7\sqrt a - 4}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{a - 5\sqrt a + 6}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{\left( {a - 2\sqrt a } \right) + \left( { - 3\sqrt a + 6} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{\sqrt a \left( {\sqrt a - 2} \right) - 3\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 4} \right)}}\\
= \dfrac{{\sqrt a - 2}}{{\sqrt a - 4}}\\
b,\\
a = 4 - 2\sqrt 3 = 3 - 2\sqrt 3 .1 + 1 = {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \sqrt a = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} = \left| {\sqrt 3 - 1} \right| = \sqrt 3 - 1\\
\Rightarrow P = \dfrac{{\left( {\sqrt 3 - 1} \right) - 2}}{{\left( {\sqrt 3 - 1} \right) - 4}} = \dfrac{{\sqrt 3 - 3}}{{\sqrt 3 - 5}} = \dfrac{{\left( {\sqrt 3 - 3} \right)\left( {\sqrt 3 + 5} \right)}}{{\left( {\sqrt 3 - 5} \right)\left( {\sqrt 3 + 5} \right)}}\\
= \dfrac{{3 + 5\sqrt 3 - 3\sqrt 3 - 15}}{{{{\sqrt 3 }^2} - {5^2}}} = \dfrac{{ - 12 + 2\sqrt 3 }}{{3 - 25}}\\
= \dfrac{{ - 12 + 2\sqrt 3 }}{{ - 22}} = \dfrac{{6 - \sqrt 3 }}{{11}}\\
c,\\
P = \dfrac{{\sqrt a - 2}}{{\sqrt a - 4}} = \dfrac{{\left( {\sqrt a - 4} \right) + 2}}{{\sqrt a - 4}} = 1 + \dfrac{2}{{\sqrt a - 4}}\\
P \in Z \Leftrightarrow \dfrac{2}{{\sqrt a - 4}} \in Z\\
a \in Z \Rightarrow \left( {\sqrt a - 4} \right) \in U\left( 2 \right) = \left\{ { - 2; - 1;1;2} \right\}\\
\Rightarrow \sqrt a \in \left\{ {2;3;5;6} \right\}\\
\Rightarrow a \in \left\{ {4;9;25;36} \right\}\\
a \ge 0,\,\,a \ne 9,\,\,a \ne 16 \Rightarrow a \in \left\{ {4;25;36} \right\}
\end{array}\)
