\(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2019
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=8\)
\(A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(A^2=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\) (do \(A>0\))
\(C=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}-\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{3}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}-\frac{\sqrt{3}-\sqrt{2}}{3}\)
\(=\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{3}=\frac{\sqrt{2}}{2}\)
VT
30 tháng 9 2019
a)=\(\sqrt{3-\sqrt{5}}\).\(\sqrt{3+\sqrt{5}}\).\(\sqrt{2}\)(\(\sqrt{5}\)-\(1\))\(\sqrt{3+\sqrt{5}}\)=2\(\sqrt{2}\) \(\sqrt{\left(\sqrt{5}-1\right)^2.\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\) .\(\sqrt{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\)\(\sqrt{8}\) =8
b)A2=8+2 căn[\(\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)\)]=8+2\(\sqrt{6-2\sqrt{5}}\)=8+2(\(\sqrt{5}\)-1)=6+2\(\sqrt{5}\)=(\(\sqrt{5}+1\))2 =>A=\(\sqrt{5}\)+1
c)C=\(\frac{2\sqrt{3}}{6}\)+\(\frac{\sqrt{2}}{6}\)-\(\frac{2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{6}\)=\(\frac{2\sqrt{3}+\sqrt{2}-2\left(\sqrt{3}-\sqrt{2}\right)}{6}\)=\(\frac{3\sqrt{2}}{6}\)=\(\frac{1}{\sqrt{2}}\)
\(A=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\sqrt{2}A=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(\sqrt{2}A=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\frac{2}{\sqrt{2}}\)
\(A=\frac{2}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}=1\)
\(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{2\left[\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)\right]}{2\left(\sqrt{10}+\sqrt{2}\right)}\)
\(=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{10}+2\sqrt{2}}\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)}{2\sqrt{10}+2\sqrt{2}}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(5+2\sqrt{5}+1\right)}{2\sqrt{10}+2\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)^2}{2\sqrt{10}+2\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)^2}{2\sqrt{10}+2\sqrt{2}}=\frac{\left(5-1\right)\left(\sqrt{5}+1\right)}{2\sqrt{10}+2\sqrt{2}}\)
\(=\frac{4\sqrt{5}+4}{2\sqrt{10}+2\sqrt{2}}=\frac{2\sqrt{5}+2}{\sqrt{10}+\sqrt{2}}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+1}\)