Question

Упростите выражение √(34-6√(3+√(13+√(48))))

Answer 1

Ответ:

$\sqrt{34-6\sqrt{3+\sqrt{13+\sqrt{48}}}}=\sqrt{34-6\sqrt{3+\sqrt{(1+\sqrt{12})^2}}}=\\\\\\=\sqrt{34-6\sqrt{3+|1+\sqrt{12}|}}=\sqrt{34-6\sqrt{3+(1+\sqrt{12})}}=\sqrt{34-6\sqrt{4+\sqrt{12}}}}=\\\\=\sqrt{34-6\sqrt{(1+\sqrt3)^2}}=\sqrt{34-6\cdot |1+\sqrt3|}=\sqrt{34-6\cdot (1+\sqrt3)}=\\\\\\=\sqrt{28-6\sqrt3}=\sqrt{27}-1=\boxed{\ 3\sqrt3-1\ }$

$\star \ \ 13+\sqrt{48}=13+\sqrt{4\cdot 12}=1+12+2\sqrt{1\cdot 12}=1^2+2\cdot 1\cdot \sqrt{12}+(\sqrt{12})^2=\\\\=(1+\sqrt{12})^2\ \ \star \\\\\star \ \ 4+\sqrt{12}=4+\sqrt{4\cdot 3}=4+2\sqrt3=1+3+2\cdot 1\cdot \sqrt3=1+2\cdot 1\cdot \sqrt3+(\sqrt3)^2=\\\\=(1+\sqrt3)^2\ \ \star$

$\star \ \ \sqrt{28-6\sqrt3}=\sqrt{28-2\cdot 3\sqrt3}=\sqrt{28-2\sqrt{27}}=\sqrt{27+1-2\sqrt{27}}=\\\\=\sqrt{(\sqrt{27})^2-2\cdot 1\cdot \sqrt{27}+1^2}=\sqrt{(\sqrt{27}-1)^2}=|\sqrt{27}-1|=3\sqrt{3}-1\ \ \star$