освободитесь от иррациональности в знаменателе дроби 2b/√a √5-√3/√5+√3. Created by rustemoksi. algebra-ru

√5+√3...

November 2021 1 14 Report

освободитесь от иррациональности в знаменателе дроби 2b/√a √5-√3/√5+√3

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bbbapho

$\frac{2b}{ \sqrt{a} } = \frac{2b}{ \sqrt{a} } \times \frac{ \sqrt{a} }{ \sqrt{a} } = \frac{2b \: \sqrt{a} }{ \sqrt{a} \: \sqrt{a} } = \frac{2b \: \sqrt{a} }{a }$

$\frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } = \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = \frac{( \sqrt{5} - \sqrt{3})( \sqrt{5} - \sqrt{3} ) }{( \sqrt{5} + \sqrt{3} )( \sqrt{5} - \sqrt{3} ) } = \frac{ {( \sqrt{5} )}^{2} - 2 \sqrt{5} \sqrt{3} + {( \sqrt{3} )}^{2} }{ {( \sqrt{5} )}^{2} - {( \sqrt{3} )}^{2} } = \frac{5 - 2 \sqrt{15} + 3}{5 - 3} = \frac{8 - 2 \sqrt{15} }{2} = \frac{2(4 - \sqrt{15} )}{2} = 4 - \sqrt{15}$

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