Question

Помогите решить с объяснением​

Answer 1

$1)\ \sqrt{(\sqrt7-2)^2}}+\sqrt{(\sqrt{7}-5)^2}=|\underbrace {\sqrt7-2}_{>0}|+|\underbrace {\sqrt7-5}_{<0}|=\\\\=(\sqrt7-2)+(5-\sqrt7)=-2+5=3$

$2)\ \ \sqrt{(\sqrt{15}-2)^2}-\sqrt{(\sqrt{15}-3)^2}=|\underbrace {\sqrt{15}-2}_{>0}|-|\underbrace{\sqrt{15}-3}_{>0}|=\\\\=(\sqrt{15}-2)-(\sqrt{15}-3)=-2+3=1$

$3)\ \ \sqrt{\sqrt{28-16\sqrt3}}=\sqrt{\sqrt{28-2\cdot \sqrt{192}}}=\sqrt{\sqrt{\underbrace {16+12}_{28}-2\sqrt{16\cdot 12}}}=\\\\\\=\sqrt{\sqrt{(\sqrt{16}-\sqrt{12})^2}}=\sqrt{\sqrt{(4-\sqrt{12})^2}}=\sqrt{|\underbrace {\, 4-\sqrt{12}\, }_{>0}|}=\sqrt{4-\sqrt{12}}=\\\\\\=\sqrt{4-2\sqrt3}=\sqrt{3+1-2\sqrt{1\cdot 3}}=\sqrt{(\sqrt3-1)^2\, }=|\, \underbrace{\sqrt3-1}_{>0}\, |=\sqrt3-1$

$4)\ \ \sqrt{\sqrt{68-\sqrt{4608}}}=\sqrt{\sqrt{68-2\sqrt{1152}}}=\sqrt{\sqrt{36+32-2\sqrt{36\cdot 32}}}=\\\\\\=\sqrt{\sqrt{(\sqrt{36}-\sqrt{32})^2}}=\sqrt{|\, 6-\sqrt{32}\, |}=\sqrt{6-\sqrt{32}}=\sqrt{6-2\sqrt{4\cdot 2}}=\\\\\\=\sqrt{4+2-2\sqrt{4\cdot 2}}=\sqrt{(\sqrt4-\sqrt2)^2}=\sqrt{(2-\sqrt2)^2}=|\, 2-\sqrt2\, |=2-\sqrt2$