## Giúp mình bài này với :v

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Giúp mình bài này với :v

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11 months 2020-11-29T00:01:37+00:00 1 Answers 57 views 0

$$\begin{array}{l} a,\\ DKXD:\,\,\,\left\{ \begin{array}{l} x \ge 0\\ y \ge 0\\ \sqrt x + \sqrt y \ne 0\\ \sqrt {xy} + y \ne 0\\ \sqrt {xy} – x \ne 0\\ \sqrt {xy} \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > 0\\ y > 0\\ x \ne y \end{array} \right.\\ M = \left( {\dfrac{{y – \sqrt {xy} }}{{\sqrt x + \sqrt y }} + \sqrt x } \right):\left( {\dfrac{x}{{\sqrt {xy} + y}} + \dfrac{y}{{\sqrt {xy} – x}} – \dfrac{{x + y}}{{\sqrt {xy} }}} \right)\\ = \dfrac{{y – \sqrt {xy} + \sqrt x \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }}:\left( {\dfrac{x}{{\sqrt y \left( {\sqrt x + \sqrt y } \right)}} – \dfrac{y}{{\sqrt x \left( {\sqrt x – \sqrt y } \right)}} – \dfrac{{x + y}}{{\sqrt {xy} }}} \right)\\ = \dfrac{{y – \sqrt {xy} + x + \sqrt {xy} }}{{\sqrt x + \sqrt y }}:\dfrac{{x\sqrt x .\left( {\sqrt x – \sqrt y } \right) – y.\sqrt y .\left( {\sqrt x + \sqrt y } \right) – \left( {x + y} \right)\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt {xy} \left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}\\ = \dfrac{{y + x}}{{\sqrt x + \sqrt y }}:\dfrac{{{x^2} – x\sqrt {xy} – y\sqrt {yx} – {y^2} – \left( {x + y} \right)\left( {x – y} \right)}}{{\sqrt {xy} \left( {x – y} \right)}}\\ = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}:\dfrac{{{x^2} – \sqrt {xy} \left( {\sqrt x + \sqrt y } \right) – {y^2} – {x^2} + {y^2}}}{{\sqrt {xy} \left( {x – y} \right)}}\\ = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}:\dfrac{{ – \sqrt {xy} \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt {xy} \left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}\\ = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}:\dfrac{{ – 1}}{{\sqrt x – \sqrt y }}\\ = \dfrac{{x + y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt x – \sqrt y }}{{ – 1}}\\ = \dfrac{{\left( {x + y} \right)\left( {\sqrt y – \sqrt x } \right)}}{{\sqrt x + \sqrt y }}\\ b,\\ y = \sqrt {1 + {8^2} + \dfrac{{{8^2}}}{{{9^2}}}} + \dfrac{8}{9} = \left( {1 + 8 – \dfrac{8}{9}} \right) + \dfrac{8}{9} = 1 + 8 = 9\\ P = \dfrac{{\left( {x + y} \right)\left( {\sqrt y – \sqrt x } \right)}}{{\sqrt x + \sqrt y }} = \dfrac{{\left( {1 + 9} \right)\left( {3 – 1} \right)}}{{3 + 1}} = \dfrac{{10.2}}{4} = 5 \end{array}$$