Đáp án:
a. \(\dfrac{{x + 1}}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
P = \dfrac{x}{{x + \sqrt x + 1}} + \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{x}{{x + \sqrt x + 1}} + \dfrac{1}{{\sqrt x - 1}} - \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x\left( {\sqrt x - 1} \right) + x + \sqrt x + 1 - x - 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - x + x + \sqrt x + 1 - x - 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - x + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 1}}{{x + \sqrt x + 1}}\\
b.P > \dfrac{2}{3}\\
\to \dfrac{{x + 1}}{{x + \sqrt x + 1}} > \dfrac{2}{3}\\
\to \dfrac{{3x + 3 - 2x - 2\sqrt x - 2}}{{3\left( {x + \sqrt x + 1} \right)}} > 0\\
\to \dfrac{{x - 2\sqrt x + 1}}{{3\left( {x + \sqrt x + 1} \right)}} > 0\\
\to \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{3\left( {x + \sqrt x + 1} \right)}} > 0\left( 1 \right)\\
Do:x + \sqrt x + 1 > 0\left( {ld} \right)\forall x \in R\\
\left( 1 \right) \to {\left( {\sqrt x - 1} \right)^2} > 0\\
\Leftrightarrow \sqrt x - 1 \ne 0\\
\Leftrightarrow x \ne 1\\
KL:x \ge 0;x \ne 1
\end{array}\)
