Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
M = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \left( {1 + \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}} \right):\dfrac{b}{{a - \sqrt {{a^2} - {b^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
{a^2} > {b^2}\\
b \ne 0
\end{array} \right)\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{\sqrt {{a^2} - {b^2}} + a}}{{\sqrt {{a^2} - {b^2}} }}:\dfrac{b}{{a - \sqrt {{a^2} - {b^2}} }}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{\left( {\sqrt {{a^2} - {b^2}} + a} \right).\left( {a - \sqrt {{a^2} - {b^2}} } \right)}}{{b.\sqrt {{a^2} - {b^2}} }}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{a^2} - \left( {{a^2} - {b^2}} \right)}}{{b.\sqrt {{a^2} - {b^2}} }}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{b^2}}}{{b.\sqrt {{a^2} - {b^2}} }}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{b}{{\sqrt {{a^2} - {b^2}} }}\\
= \dfrac{{a - b}}{{\sqrt {{a^2} - {b^2}} }}\\
5,\\
P = \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right)\\
= \left( {\dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left[ {\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)} \right]}^2}}}{2}\\
= \dfrac{{\left( {x - \sqrt x - 2} \right) - \left( {x + \sqrt x - 2} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\
= \dfrac{{ - 2\sqrt x }}{1}.\dfrac{{\sqrt x - 1}}{2}\\
= - \sqrt x \left( {\sqrt x - 1} \right)
\end{array}\)
