Đáp án:
d. \(\dfrac{7}{{\sqrt x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.N = \left[ {\dfrac{{\sqrt x + 1 - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right].\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x + 1}}\\
= \dfrac{2}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x + 1}}\\
= \dfrac{2}{{x + 1}}\\
b.P = \left[ {\dfrac{{1 - a\sqrt a + \sqrt a - a}}{{1 - \sqrt a }}} \right]{\left( {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right)^2}\\
= \left[ {\dfrac{{\left( {1 - \sqrt a } \right){{\left( {\sqrt a + 1} \right)}^2}}}{{1 - \sqrt a }}} \right]{\left[ {\dfrac{1}{{\left( {1 + \sqrt a } \right)}}} \right]^2}\\
= {\left( {\sqrt a + 1} \right)^2}.\dfrac{1}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= 1\\
c.H = \dfrac{{x\sqrt x - y\sqrt y + x\sqrt y - y\sqrt x }}{{\sqrt x - \sqrt y }}.{\left[ {\dfrac{{\sqrt x - \sqrt y }}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right]^2}\\
= \dfrac{{x\left( {\sqrt x + \sqrt y } \right) - y\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x - \sqrt y }}.\left[ {\dfrac{{\sqrt x - \sqrt y }}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right]\\
= \dfrac{{\left( {x - y} \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x - \sqrt y }}.\left[ {\dfrac{{\sqrt x - \sqrt y }}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right]\\
= \sqrt x + \sqrt y \\
d.Q = \dfrac{{3\sqrt x + 6 + 4\sqrt x - 8 - 12}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{7\sqrt x - 14}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{7\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{7}{{\sqrt x + 2}}
\end{array}\)