Đáp án:
\(\begin{array}{l}
9,\\
P = \dfrac{6}{{\sqrt x }}\\
10,\\
P = \dfrac{{ - \sqrt x }}{{\sqrt x - 2}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
9,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x + 1 \ne 0\\
\sqrt x - 1 \ne 0\\
1 - x \ne 0\\
\sqrt x \ne 0\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
P = \left( {\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} + \dfrac{{2\sqrt x }}{{\sqrt x - 1}} - \dfrac{{3\sqrt x - 1}}{{1 - x}}} \right).\left( {\dfrac{2}{{\sqrt x }} - \dfrac{2}{x}} \right)\\
= \left( {\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} + \dfrac{{2\sqrt x }}{{\sqrt x - 1}} + \dfrac{{3\sqrt x - 1}}{{x - 1}}} \right).\left( {\dfrac{2}{{\sqrt x }} - \dfrac{2}{x}} \right)\\
= \left( {\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} + \dfrac{{2\sqrt x }}{{\sqrt x - 1}} + \dfrac{{3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right).\left( {\dfrac{2}{{\sqrt x }} - \dfrac{2}{x}} \right)\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2} + 2\sqrt x \left( {\sqrt x + 1} \right) + \left( {3\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{2\sqrt x - 2}}{x}\\
= \dfrac{{x - 2\sqrt x + 1 + 2x + 2\sqrt x + 3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{2\left( {\sqrt x - 1} \right)}}{x}\\
= \dfrac{{3x + 3\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{2\left( {\sqrt x - 1} \right)}}{x}\\
= \dfrac{{3\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{2\left( {\sqrt x - 1} \right)}}{x}\\
= \dfrac{{3\sqrt x .2}}{x}\\
= \dfrac{6}{{\sqrt x }}\\
10,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x + 2 \ne 0\\
\sqrt x - 2 \ne 0\\
4 - x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} - \dfrac{{2\sqrt x }}{{\sqrt x - 2}} + \dfrac{{2 - 5\sqrt x }}{{4 - x}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} - \dfrac{{2\sqrt x }}{{\sqrt x - 2}} - \dfrac{{2 - 5\sqrt x }}{{x - 4}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} - \dfrac{{2\sqrt x }}{{\sqrt x - 2}} - \dfrac{{2 - 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) - 2\sqrt x \left( {\sqrt x + 2} \right) - \left( {2 - 5\sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {x - 3\sqrt x + 2} \right) - \left( {2x + 4\sqrt x } \right) - \left( {2 - 5\sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2 - 2x - 4\sqrt x - 2 + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{ - x - 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{ - \sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{ - \sqrt x }}{{\sqrt x - 2}}
\end{array}\)
