Giải thích các bước giải:
C2:
a) ĐKXĐ: $x>0;x\ne 1$
Ta có:
$\begin{array}{l}
P = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x }}{{x - 2\sqrt x + 1}}\\
= \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{x}\\
= \dfrac{{x - 1}}{x}\\
= 1 - \dfrac{1}{x}
\end{array}$
Vậy $P = 1 - \dfrac{1}{x}$ với $x>0;x\ne 1$
b) Ta có:
$\begin{array}{l}
P > \dfrac{1}{2}\\
\Leftrightarrow 1 - \dfrac{1}{x} > \dfrac{1}{2}\\
\Leftrightarrow \dfrac{1}{x} < \dfrac{1}{2}\\
\Leftrightarrow x < 2\left( {do:x > 0} \right)
\end{array}$
Vậy $x<2$ thỏa mãn.
C5:
ĐK: $a,b\le 0$
Ta có:
$\begin{array}{l}
P = \dfrac{1}{a} + \dfrac{1}{b}\\
= \dfrac{{{1^2}}}{a} + \dfrac{{{1^2}}}{b}\\
\ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{a + b}}\\
= \dfrac{4}{{a + b}}\\
\ge \dfrac{4}{{2\sqrt 2 }}\\
= \sqrt 2
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{a} = \dfrac{1}{b}\\
a + b = 2\sqrt 2
\end{array} \right.\\
\Leftrightarrow a = b = \sqrt 2
\end{array}$
Vậy $MinP = \sqrt 2 \Leftrightarrow a = b = \sqrt 2 $
