## Giúp em câu c,d với ạ. Em đang cần gấp chiều nay em đi học rồi ạ

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Giúp em câu c,d với ạ. Em đang cần gấp chiều nay em đi học rồi ạ

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12 months 2020-11-08T00:43:39+00:00 1 Answers 55 views 0

$$\begin{array}{l} A = \dfrac{{2\sqrt x }}{{\sqrt x – 3}} – \dfrac{{x + 9\sqrt x }}{{x – 9}}\\ = \dfrac{{2\sqrt x }}{{\sqrt x – 3}} – \dfrac{{x + 9\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) – \left( {x + 9\sqrt x } \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{\left( {2x + 6\sqrt x } \right) – \left( {x + 9\sqrt x } \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{x – 3\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\ B = \dfrac{{x – 5\sqrt x }}{{x – 25}} = \dfrac{{\sqrt x \left( {\sqrt x – 5} \right)}}{{\left( {\sqrt x – 5} \right)\left( {\sqrt x + 5} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x + 5}}\\ M = \dfrac{A}{B} = \dfrac{{\sqrt x }}{{\sqrt x + 3}}:\dfrac{{\sqrt x }}{{\sqrt x + 5}} = \dfrac{{\sqrt x + 5}}{{\sqrt x + 3}}\\ M – 1 = \dfrac{{\sqrt x + 5}}{{\sqrt x + 3}} – 1 = \dfrac{{\left( {\sqrt x + 5} \right) – \left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}} = \dfrac{2}{{\sqrt x + 3}} > 0\\ \Rightarrow M > 1\\ d,\\ B = \dfrac{{\sqrt x }}{{\sqrt x + 5}} = \dfrac{{\left( {\sqrt x + 5} \right) – 5}}{{\sqrt x + 5}} = 1 – \dfrac{5}{{\sqrt x + 5}}\\ B \in Z \Rightarrow \dfrac{5}{{\sqrt x + 5}} \in Z \Rightarrow \left( {\sqrt x + 5} \right) \in \left\{ { \pm 1; \pm 5} \right\}\\ \sqrt x \ge 0 \Rightarrow \sqrt x + 5 \ge 5 \Rightarrow \sqrt x + 5 = 5 \Rightarrow x = 0 \end{array}$$