Упростите! [tex]( \frac{ \sqrt{a} }{ \sqrt{a} - \sqrt{b} } - \frac{ \sqrt{b} }{ \sqrt{a} + \sqrt{b} }) \frac{a-b}{ a^{2}+ab } [/tex]. Created by Yana4660. algebra-ru

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Упростите!
[tex]( \frac{ \sqrt{a} }{ \sqrt{a} - \sqrt{b} } - \frac{ \sqrt{b} }{ \sqrt{a} + \sqrt{b} }) \frac{a-b}{ a^{2}+ab } [/tex]

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kmike21

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$( \frac{ \sqrt{a} }{ \sqrt{a} - \sqrt{b} } - \frac{ \sqrt{b} }{ \sqrt{a} + \sqrt{b} }) \frac{a-b}{ a^{2}+ab } = ( \frac{ \sqrt{a} }{ \sqrt{a} - \sqrt{b} } - \frac{ \sqrt{b} }{ \sqrt{a} + \sqrt{b} }) \frac{( \sqrt{a}-\sqrt{b})( \sqrt{a}+\sqrt{b})}{ a(a+b) }=$
$( \frac{ \sqrt{a}( \sqrt{a}-\sqrt{b})( \sqrt{a}+\sqrt{b}) }{ \sqrt{a} - \sqrt{b} } - \frac{ \sqrt{b}( \sqrt{a}-\sqrt{b})( \sqrt{a}+\sqrt{b}) }{ \sqrt{a} + \sqrt{b} }) \frac{1}{ a(a+b) }=$
$( \sqrt{a}( \sqrt{a}+\sqrt{b}) - \sqrt{b}( \sqrt{a}-\sqrt{b})) \frac{1}{ a(a+b) }=$
$\frac{( \sqrt{a}( \sqrt{a}+\sqrt{b}) - \sqrt{b}( \sqrt{a}-\sqrt{b})) }{ a(a+b) }=\frac{(a+\sqrt{ab}) - ( \sqrt{ab}-b) }{ a(a+b) }=\frac{a+\sqrt{ab} - \sqrt{ab}+b }{ a(a+b) }=\frac{a+b }{ a(a+b) }=$
$=\frac{1}{a}$

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