Предмет: Алгебра, автор: nili06

Вычислить: $\sqrt{34+24 \sqrt{2} } + \sqrt{34-24 \sqrt{2} }$

Ответы

Автор ответа: NNNLLL54

$\sqrt{34+24 \sqrt{2}}+\sqrt{34-24\sqrt2} =\\\\=\sqrt{(\sqrt{18}+\sqrt{16})^2}+\sqrt{(\sqrt{18}-\sqrt{16})^2}=\\\\=|\sqrt{18}+\sqrt{16}|+|\sqrt{18}- \sqrt{16} |=\\\\=\sqrt{18}+ \sqrt{16}+\sqrt{18} - \sqrt{16}=2 \sqrt{18}=2\sqrt{9\cdot 2}=6\sqrt2$

Автор ответа: kolobok1431

$\sqrt{ (\sqrt{34+24 \sqrt{2} } + \sqrt{34-24 \sqrt{2} } )^{2}} =$
$= \sqrt{(34+24+2 \sqrt{(34+24 \sqrt{2} )*(34-24 \sqrt{2}) }+34-24 \sqrt{2} }=$ $\sqrt{68+2 \sqrt{ 34^{2}- 24^{2}*2 } } = \sqrt{68+ 2\sqrt{1156-576*2} }= \sqrt{68+2 \sqrt{4} } =$ $\sqrt{68+2*2}= \sqrt{72}= \sqrt{36*2} =6 \sqrt{2}$
Данное выражение возвели в квадрат, упростили, а потом из него извлекли квадратный корень

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