Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
P = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x > 0\\
x \ne 1\\
x \ne 4
\end{array} \right)\\
= \dfrac{{\sqrt x - \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right).\sqrt x }}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {x - 1} \right) - \left( {x - 4} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{3}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{3}\\
= \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
b,\\
P > 0 \Leftrightarrow \dfrac{{\sqrt x - 2}}{{3\sqrt x }} > 0 \Leftrightarrow \sqrt x - 2 > 0 \Leftrightarrow x > 4\\
3,\\
x - \sqrt {x - 7} = 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 7} \right)\\
\Leftrightarrow x - \sqrt {x - 7} - 9 = 0\\
\Leftrightarrow \left( {x - 7} \right) - \sqrt {x - 7} - 2 = 0\\
\Leftrightarrow {\sqrt {x - 7} ^2} - \sqrt {x - 7} - 2 = 0\\
\Leftrightarrow \left( {\sqrt {x - 7} - 2} \right)\left( {\sqrt {x - 7} + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 7} = 2\\
\sqrt {x - 7} = - 1\,\,\,\,\,\left( L \right)
\end{array} \right. \Rightarrow \sqrt {x - 7} = 2 \Leftrightarrow x = 11
\end{array}\)
