Đáp án:
\(\begin{array}{l}
a) - \dfrac{3}{{\sqrt x + 1}}\\
b)\dfrac{2}{{\sqrt x + 1}}\\
c) - \dfrac{3}{{\sqrt x + 1}}\\
d)\dfrac{2}{{\sqrt x + 1}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{\sqrt x \left( {\sqrt x - 1} \right) - \sqrt x \left( {\sqrt x + 1} \right) + 3 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x - x - \sqrt x + 3 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 3\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= - \dfrac{3}{{\sqrt x + 1}}\\
b)\dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2} - 4}}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1 - 4}}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{4\sqrt x - 4}}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{2}{{\sqrt x + 1}}\\
c)\dfrac{{\sqrt x \left( {\sqrt x - 1} \right) - \sqrt x \left( {\sqrt x + 1} \right) + 3 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x - x - \sqrt x + 3 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 3\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= - \dfrac{3}{{\sqrt x + 1}}\\
d)\left[ {\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{{{\left( {\sqrt x - 2} \right)}^2}}} - \dfrac{{\sqrt x }}{{\sqrt x - 2}}} \right].\dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 2 - \sqrt x }}{{\sqrt x - 2}}.\dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
= \dfrac{2}{{\sqrt x + 1}}
\end{array}\)
