Đáp án:
2) \(\dfrac{2}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)A = \dfrac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{1 - \sqrt x + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{x + \sqrt x + 1}}\\
2)B = \dfrac{{\sqrt x - 1}}{2}\\
P = A:B = \dfrac{{\sqrt x - 1}}{{x + \sqrt x + 1}}:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
3)\dfrac{1}{P} = \dfrac{{x + \sqrt x + 1}}{2}\\
Có:\dfrac{1}{P} > m + \sqrt x \\
\to \dfrac{{x + \sqrt x + 1}}{2} > m + \sqrt x \\
\to \dfrac{{x + \sqrt x + 1 - 2m - 2\sqrt x }}{2} > 0\\
\to x - \sqrt x + 1 - 2m > 0\forall x > 1\\
\Leftrightarrow \left( {x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{3}{4} - 2m > 0\forall x > 1\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} - 2m > 0\forall x > 1\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} > 0\forall x > 1\\
\to \dfrac{3}{4} - 2m > 0\\
\to \dfrac{3}{4} > 2m\\
\to \dfrac{3}{8} > m
\end{array}\)
