Đáp án:
1) \(A = \dfrac{3}{5}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 16\\
\to A = \dfrac{{\sqrt {16} - 1}}{{\sqrt {16} + 1}} = \dfrac{3}{5}\\
2)B = \dfrac{{x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{x - \sqrt x + 1}} - \dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{x + 2 + \sqrt x \left( {\sqrt x + 1} \right) - x + \sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 + x + \sqrt x - x + \sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{x - \sqrt x + 1}}\\
\to dpcm\\
3)A.B < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{x - \sqrt x + 1}} < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 1}}{{x - \sqrt x + 1}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - 2 - x + \sqrt x - 1}}{{2\left( {x - \sqrt x + 1} \right)}} < 0\\
\to \dfrac{{ - x + 3\sqrt x - 3}}{{2\left( {x - \sqrt x + 1} \right)}} < 0\\
Do:\left\{ \begin{array}{l}
- x + 3\sqrt x - 3 < 0\forall x \ge 0\\
x - \sqrt x + 1 > 0\forall x \ge 0
\end{array} \right.\\
\to \dfrac{{ - x + 3\sqrt x - 3}}{{2\left( {x - \sqrt x + 1} \right)}} < 0\left( {ld} \right)\\
KL:x \ge 0
\end{array}\)
