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Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Ta có \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\left(\sqrt{10}-\sqrt{2}\right)\)
= \(2\sqrt{4+\sqrt{\sqrt{5}^2-2\sqrt{5}.1+1}}\sqrt{2}\left(\sqrt{5}-1\right)\)
= \(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\sqrt{2}\left(\sqrt{5}-1\right)\)
= \(\sqrt{2}\sqrt{4+\sqrt{5}-1}.\left(\sqrt{5}-1\right)2\)
= \(\sqrt{2\left(3+\sqrt{5}\right)}\left(\sqrt{5}-1\right)2\)
= \(\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)2\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)2\)
= \(\left(\sqrt{5}^2-1\right)2\)
= 4.2
= 8
Chúc bạn làm bài tốt :)
Lời giải:
\(N=\sqrt{4\sqrt{6}+8\sqrt{3}+4\sqrt{2}+18}\)
\(=\sqrt{2\sqrt{24}+4(2\sqrt{3}+\sqrt{2})+18}\)
\(=\sqrt{12+2\sqrt{24}+2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2})^2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2}+2)^2}=\sqrt{12}+\sqrt{2}+2=2\sqrt{3}+\sqrt{2}+2\)
\(A=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
=\(\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(=3-\sqrt{3}+2\sqrt{3}-3\)
=\(\sqrt{3}\)
