Đáp án:
a) \(\dfrac{2}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Q = \left[ {\dfrac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right]:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x + 2 + x - \sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 1}}{{x + \sqrt x + 1}}.\dfrac{2}{{\sqrt x - 1}} = \dfrac{2}{{x + \sqrt x + 1}}\\
b)Do:x + \sqrt x + 1 = x + 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x \ge 0;x \ne 1\\
\to Q > 0\forall x \ge 0;x \ne 1
\end{array}\)
