Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} + \dfrac{{8\sqrt x - 4}}{{1 - x}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x - 4}}{{x - 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x - 4}}{{{{\sqrt x }^2} - {1^2}}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2} - \left( {8\sqrt x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {x + 2\sqrt x + 1} \right) - \left( {x - 2\sqrt x + 1} \right) - \left( {8\sqrt x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1 - 8\sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 4\sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 4.\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 4}}{{\sqrt x + 1}}\\
2,\\
A = \left( {\dfrac{1}{{\sqrt x - 3}} - \dfrac{1}{{\sqrt x + 3}}} \right):\dfrac{3}{{\sqrt x - 3}}\\
= \dfrac{{\left( {\sqrt x + 3} \right) - \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{3}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x + 3 - \sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{3}{{\sqrt x - 3}}\\
= \dfrac{6}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{2}{{\sqrt x + 3}}\\
3,\\
1,\\
B = \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x + 1}}{{x + \sqrt x }}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {{{\sqrt x }^2} - {1^2}} \right) + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 1 + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
2,\\
A:B = \dfrac{{\sqrt x }}{{\sqrt x + 1}}:\dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x }}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} = \dfrac{{\sqrt x }}{{\sqrt x + 2}}
\end{array}\)