## Giúp em.bài 1vs………

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Giúp em.bài 1vs………

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1 year 2020-11-10T08:18:10+00:00 2 Answers 65 views 0

$$\begin{array}{l} a,\\ DKXD:\,\,\,\left\{ \begin{array}{l} x \ge 0\\ \sqrt x – 2 \ne 0\\ \sqrt x + 2 \ne 0\\ 4 – x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 4 \end{array} \right.\\ b,\\ A = \dfrac{1}{{\sqrt x – 2}} + \dfrac{1}{{\sqrt x + 2}} – \dfrac{x}{{4 – x}}\\ = \dfrac{1}{{\sqrt x – 2}} + \dfrac{1}{{\sqrt x + 2}} – \dfrac{x}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}\\ = \dfrac{{\left( {\sqrt x + 2} \right) + \left( {\sqrt x – 2} \right) + x}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{\sqrt x }}{{\sqrt x – 2}}\\ c,\\ x = 9 \Rightarrow A = \dfrac{{\sqrt 9 }}{{\sqrt 9 – 2}} = \dfrac{3}{{3 – 2}} = 3\\ d,\\ x = 3 – 2\sqrt 2 = 2 – 2.\sqrt 2 .1 + 1 = {\left( {\sqrt 2 – 1} \right)^2}\\ \Rightarrow \sqrt x = \sqrt 2 – 1\\ \Rightarrow B = \dfrac{{\left( {\sqrt 2 – 1} \right) + 3}}{{2 – \left( {\sqrt 2 – 1} \right)}} = \dfrac{{\sqrt 2 + 2}}{{3 – \sqrt 2 }} = \dfrac{{\left( {\sqrt 2 + 2} \right)\left( {3 + \sqrt 2 } \right)}}{{\left( {3 – \sqrt 2 } \right)\left( {3 + \sqrt 2 } \right)}}\\ = \dfrac{{3\sqrt 2 + 2 + 6 + 2\sqrt 2 }}{{9 – 2}} = \dfrac{{8 + 5\sqrt 2 }}{7}\\ e,\\ P = B:A = \dfrac{{\sqrt x + 3}}{{2 – \sqrt x }}:\dfrac{{\sqrt x }}{{\sqrt x – 2}} = \dfrac{{\sqrt x + 3}}{{2 – \sqrt x }}.\dfrac{{\sqrt x – 2}}{{\sqrt x }} = – \dfrac{{\sqrt x + 3}}{{\sqrt x }}\\ P = – 2 \Leftrightarrow – \dfrac{{\sqrt x + 3}}{{\sqrt x }} = – 2\\ \Leftrightarrow \sqrt x + 3 = 2\sqrt x \\ \Leftrightarrow \sqrt x = 3\\ \Leftrightarrow x = 9\\ g,\\ \dfrac{{\sqrt x + 1}}{{\sqrt x }} > 0 \Rightarrow – \dfrac{{\sqrt x + 1}}{{\sqrt x }} < 0 \Rightarrow P < 0 \Rightarrow P < 1\\ h,\\ P < 1 \Rightarrow bpt\,\,\,P \ge 3\,\,vn\\ k,\\ A = \dfrac{{\sqrt x }}{{\sqrt x – 2}} = \dfrac{{\left( {\sqrt x – 2} \right) + 2}}{{\sqrt x – 2}} = 1 + \dfrac{2}{{\sqrt x – 2}}\\ A \in Z \Leftrightarrow \dfrac{2}{{\sqrt x – 2}} \in Z\\ x \in Z \Rightarrow \left( {\sqrt x – 2} \right) \in \left\{ { \pm 1; \pm 2} \right\}\\ \Rightarrow \sqrt x \in \left\{ {0;1;3;4} \right\}\\ \Rightarrow x \in \left\{ {0;1;9;16} \right\}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right) \end{array}$$