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July 2022 1 7 Report

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[tex]( \frac{ \sqrt{x} +3 \sqrt{y} }{ \sqrt{x} + \sqrt{y} } - ( \sqrt{x} - \sqrt{y} )( \sqrt{x} + \sqrt{y} ) ^{-1} ) * \frac{ \sqrt{x} + \sqrt{y} } {8( \sqrt{y} ) ^{3} } [/tex]

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NNNLLL54

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$\left ( \frac{ \sqrt{x}+3\sqrt{y} }{ \sqrt{x} + \sqrt{y} } -( \sqrt{x}-\sqrt{y} )( \sqrt{x}+\sqrt{y} )^{-1}\right )\cdot \frac{ \sqrt{x} +\sqrt{y} }{8(\sqrt{y})^3} =\\\\=\left ( \frac{ \sqrt{x} +3\sqrt{y} }{ \sqrt{x}+\sqrt{y} }-\frac{ \sqrt{x}-\sqrt{y} }{\sqrt{x}+\sqrt{y}} \right )\cdot \frac{\sqrt{x}+\sqrt{y}}{8(\sqrt{y})^3}=$

$= \frac{ \sqrt{x} +3 \sqrt{y}- \sqrt{x} +\sqrt{y} }{\sqrt{x}+\sqrt{y} } \cdot \frac{ \sqrt{x}+\sqrt{y} }{8y^{\frac{3}{2}}} = \frac{4 \sqrt{y} }{8y^{\frac{3}{2}}} = \frac{1}{2y}$

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MadelineByOne August 2022 | 0 Ответы

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