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0}|=\sqrt3+\sqrt2" alt="3)\; \; \sqrt{5+2\sqrt6}=\sqrt{5+2\cdot \sqrt3\cdot \sqrt2}}=\sqrt{3+2\cdot \sqrt3\cdot \sqrt2+2}=\\\\=\sqrt{(\sqrt3+\sqrt2)^2}=|\underbrace {\sqrt3+\sqrt2}_{>0}|=\sqrt3+\sqrt2" align="absmiddle" class="latex-formula">

0}|=\sqrt5-\sqrt3" alt="4)\; \; \sqrt{8-2\sqrt{15}}=\sqrt{\underline {5+3}-2\cdot \sqrt5\cdot \sqrt3}=\sqrt{(\sqrt5-\sqrt3)^2}=\\\\=|\underbrace {\sqrt5-\sqrt3}_{>0}|=\sqrt5-\sqrt3" align="absmiddle" class="latex-formula">

$1)\; \; \sqrt{5+\sqrt{21}}=\sqrt{\dfrac{10+2\sqrt{21}}{2}}=\dfrac{\sqrt{7+3+2\cdot \sqrt7\cdot \sqrt3}}{\sqrt2}=\dfrac{\sqrt{(\sqrt7+\sqrt3)^2}}{\sqrt2}=\\\\=\dfrac{|\sqrt7+\sqrt3|}{\sqrt2}=\dfrac{\sqrt7+\sqrt3}{\sqrt2}\; ;\; \; \; \; \Big(\; \dfrac{\sqrt{14}+\sqrt6}{2}\; \Big)$

$2)\; \; \sqrt{4+\sqrt7}=\sqrt{\dfrac{8+2\sqrt7}{2}}=\dfrac{\sqrt{8+2\sqrt7}}{\sqrt2}=\dfrac{\sqrt{7+1+2\cdot \sqrt7\cdot \sqrt1}}{\sqrt2}=\\\\\\=\dfrac{\sqrt{(\sqrt7+1)^2}}{\sqrt2}=\dfrac{|\sqrt7+1|}{\sqrt2}=\dfrac{\sqrt7+1}{\sqrt2}\; ;\; \; \; \; \Big (\dfrac{\sqrt{14}+\sqrt2}{2}\; \Big)$

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√(5 +√21 ) = √(10 +2√(7*3) ) /√2 =√( (√7)² +2√7*√3+(√(3)² ) / √2 =

=√( ( √7 +√3 )² ) / √2 = (√7 +√3 ) / √2 = 0,5 ( √14 +√6 ) .

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√(4 +√7 ) = √(8 +2√(7*1) ) /√2 = √( (√7)² +2√7*1+1² ) / √2 =

=√( ( √7 +1 )² ) / √2 = (√7 +1 ) / √2 = 0,5 ( √14 +√2 ) .

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√(5 +2√6 ) = √(3 +2√(3*2) +2 ) = √( (√3)² +2√3*√2+(√2)² ) =

=√( ( √3 +√2 )² ) = √3 +√2 .

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√(8 -2√15 ) = √(5 - 2√(5*3) +3 ) = √( (√5)² - 2√5*√3+(√3) ² ) =

=√( ( √5 - √3 )² ) = √5 - √3 .

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