Đáp án:
$P = \dfrac{{\sqrt x - 3}}{{\sqrt x - 1}},x \ge 0;x \ne 1$
Giải thích các bước giải:
ĐKXĐ: $x \ge 0;x \ne 1$
Ta có:
$\begin{array}{l}
P = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{x\sqrt x - \sqrt x + x - 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x - 1}}} \right)\\
= \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\left( {\sqrt x + 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{x - 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right):\left( {\dfrac{{\sqrt x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \left( {\dfrac{{\sqrt x - 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\left( {\dfrac{{\sqrt x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{\sqrt x - 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 1}}
\end{array}$
Vậy $P = \dfrac{{\sqrt x - 3}}{{\sqrt x - 1}},x \ge 0;x \ne 1$
