## Giải giúp mình bài 2 với ạ, mình cảm ơn

Question

Giải giúp mình bài 2 với ạ, mình cảm ơn

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1 year 2020-10-22T12:27:19+00:00 1 Answers 58 views 0

## Answers ( )

1. Đáp án:

$\begin{array}{l} a)Dkxd:\left\{ \begin{array}{l} x \ge 0\\ x – 2\sqrt x \ne 0\\ \sqrt x – 2 \ne 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} x \ge 0\\ \sqrt x \left( {\sqrt x – 2} \right) \ne 0\\ \sqrt x – 2 \ne 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} x > 0\\ x \ne 4 \end{array} \right.\\ A = \left( {\dfrac{{\sqrt x + 1}}{{x – 2\sqrt x }} – \dfrac{1}{{\sqrt x – 2}}} \right).\left( {x – 3\sqrt x + 2} \right)\\ = \left( {\dfrac{{\sqrt x + 1 – \sqrt x }}{{\sqrt x \left( {\sqrt x – 2} \right)}}} \right).\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)\\ = \dfrac{1}{{\sqrt x \left( {\sqrt x – 2} \right)}}.\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)\\ = \dfrac{{\sqrt x – 1}}{{\sqrt x }}\\ b)A < \dfrac{1}{2}\\ \Rightarrow \dfrac{{\sqrt x – 1}}{{\sqrt x }} < \dfrac{1}{2}\\ \Rightarrow \dfrac{{\sqrt x – 1}}{{\sqrt x }} – \dfrac{1}{2} < 0\\ \Rightarrow \dfrac{{2\sqrt x – 2 – \sqrt x }}{{2\sqrt x }} < 0\\ \Rightarrow \sqrt x – 2 < 0\left( {do:\sqrt x > 0} \right)\\ \Rightarrow \sqrt x < 2\\ \Rightarrow x < 4\\ Vay\,0 < x < 4\\ c)A = \dfrac{{\sqrt x – 1}}{{\sqrt x }} = 1 – \dfrac{1}{{\sqrt x }}\\ A \in Z\\ \Rightarrow \dfrac{1}{{\sqrt x }} \in Z\\ \Rightarrow \sqrt x = 1\left( {do:\sqrt x > 0} \right)\\ \Rightarrow x = 1\left( {tmdk} \right)\\ Vay\,x = 1 \end{array}$