106 плесссссссссссссссс

Решите задачу:

$1)\; \sqrt{3-2\sqrt2} =\sqrt{1+2-2\sqrt2}=\sqrt{(1-\sqrt2)^2}=|1-\sqrt2|=\sqrt2-1\\\\\star \; 1-\sqrt2\approx 1-1,4=-0,4\ \textless \ 0\; \; \to \; \; |1-\sqrt2|=-(1-\sqrt2)=\sqrt2-1\\\\2)\; \sqrt{4+2\sqrt3}=\sqrt{1+3+2\sqrt3}=\sqrt{(1+\sqrt3)^2}=|1+\sqrt3|=1+\sqrt3\\\\3)\; \sqrt{7+2\sqrt6}=\sqrt{1+6+2\sqrt6}=\sqrt{(1+\sqrt6)^2}=|1+\sqrt6|=1+\sqrt6\\\\4)\; \sqrt{11-2\sqrt{10}}=\sqrt{1+10-2\sqrt{10}}=\sqrt{(1-\sqrt{10})^2}=|1-\sqrt{10}|=\\\\=-(1-\sqrt{10})=\sqrt{10}-1$

$\star\; \; (1-\sqrt{10})\ \textless \ 0\; \; \to \; \; |1-\sqrt{10}|=-(1-\sqrt{10})=\sqrt{10}-1$