Question

Решите №10

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.20 баллов

Answer 1

Объяснение:

$\sqrt{34-6\sqrt{3+\sqrt{13+\sqrt{48} } } }.\\1)\\\sqrt{13+\sqrt{48} } =\sqrt{12+1+\sqrt{4*12} } =\sqrt{(\sqrt{12})^2+2\sqrt{12}*1+1^2} }}=\\=\sqrt{(\sqrt{12} +1)^2}=|\sqrt{12}+1|= \sqrt{12}+1.$

$2)\\6*\sqrt{3+\sqrt{12} +1} =6*\sqrt{4+\sqrt{12} }=6*\sqrt{3+1+\sqrt{4*3} } =\\=6*\sqrt{(\sqrt{3)^2}+2\sqrt{3}*1+1^2 } =6*\sqrt{(\sqrt{3}+1 )^2}=6*|\sqrt{3}+1|=6{\sqrt{3} +6.}$

$3)\\\sqrt{34-(6\sqrt{3}+6) } =\sqrt{34-6\sqrt{3}-6 } =\sqrt{28-6\sqrt{3} } =\sqrt{27+1-6\sqrt{3} } =\\=\sqrt{(\sqrt{27})^2-2*3\sqrt{3}+1 } =\sqrt{(3\sqrt{3})^2 -2*3\sqrt{3}*1+1^2 }=\\ =\sqrt{(3\sqrt{3}-1)^2 } =|3\sqrt{3} -1|=3\sqrt{3}-1.$

Ответ: 3√3-1.