Câu hỏi của Huỳnh Võ Niên Tường

Question

$\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{12}\right).\left(2+\sqrt{3}\right)$

$\sqrt{2}.\sqrt{2-\sqrt{3}}.\left(\sqrt{3}+1\right)$

$\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}$

\(\sqrt{\sqrt{5}-\sq...

Nguyễn Việt Lâm · Accepted Answer

$=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)$

$=\sqrt{\left(\sqrt{3}-1\right)^2}\left(3-4\right)$

$=\left(\sqrt{3}-1\right).\left(-1\right)=1-\sqrt{3}$

b/ $=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)$

$=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2$

c/ $=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}$

$=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=\sqrt{9}=3$

d/ $=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}$

$=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}$

$=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1$Câu trả lời: $=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)$