## giúp mik kkkkkkkkkkkkkkkkkkkkkkk

Question

giúp mik kkkkkkkkkkkkkkkkkkkkkkk

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4 years 2020-11-11T03:55:01+00:00 2 Answers 72 views 0

1. Đáp án:

1) $$A = \dfrac{3}{5}$$

Giải thích các bước giải:

$$\begin{array}{l} 1)Thay:x = 16\\ \to A = \dfrac{{\sqrt {16} – 1}}{{\sqrt {16} + 1}} = \dfrac{3}{5}\\ 2)B = \dfrac{{x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{x – \sqrt x + 1}} – \dfrac{1}{{\sqrt x + 1}}\\ = \dfrac{{x + 2 + \sqrt x \left( {\sqrt x + 1} \right) – x + \sqrt x – 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\ = \dfrac{{x + 2 + x + \sqrt x – x + \sqrt x – 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\ = \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\ = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\ = \dfrac{{\sqrt x + 1}}{{x – \sqrt x + 1}}\\ \to dpcm\\ 3)A.B < \dfrac{1}{2}\\ \to \dfrac{{\sqrt x – 1}}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{x – \sqrt x + 1}} < \dfrac{1}{2}\\ \to \dfrac{{\sqrt x – 1}}{{x – \sqrt x + 1}} < \dfrac{1}{2}\\ \to \dfrac{{2\sqrt x – 2 – x + \sqrt x – 1}}{{2\left( {x – \sqrt x + 1} \right)}} < 0\\ \to \dfrac{{ – x + 3\sqrt x – 3}}{{2\left( {x – \sqrt x + 1} \right)}} < 0\\ Do:\left\{ \begin{array}{l} – x + 3\sqrt x – 3 < 0\forall x \ge 0\\ x – \sqrt x + 1 > 0\forall x \ge 0 \end{array} \right.\\ \to \dfrac{{ – x + 3\sqrt x – 3}}{{2\left( {x – \sqrt x + 1} \right)}} < 0\left( {ld} \right)\\ KL:x \ge 0 \end{array}$$

2. Đáp án:

1) A

Giải thích các bước giải: