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NH
1
6 tháng 6 2019
a) \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\sqrt{10}+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\sqrt{10}-1\)
b) \(\sqrt{21-6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}-\sqrt{3}\)
VI
1
7 tháng 9 2016
a)\(\sqrt{1-2\sqrt{10}+10}=\sqrt{\left(1-\sqrt{10}\right)^2}=\left|1-\sqrt{10}\right|=\sqrt{10}-1\)
(vì 1<\(\sqrt{10}\))
b)\(\Rightarrow\sqrt{2}\left[\left(\sqrt{4-\sqrt{7}}\right)-\left(\sqrt{4+\sqrt{7}}\right)\right]=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}=\sqrt{7}-1-1-\sqrt{7}=-2\Rightarrow\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
25 tháng 10 2015
A=\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{5}+1-\sqrt{7+2\sqrt{10}}}\)=\(\frac{\sqrt{2}\left(\sqrt{3}+3+\sqrt{2}-\sqrt{5+2\sqrt{6}}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{5}+1-\sqrt{7+2\sqrt{10}}\right)}\)
A=\(\frac{\sqrt{6}+3\sqrt{2}+2-\sqrt{10+4\sqrt{6}}}{2+\sqrt{10}+\sqrt{2}-\sqrt{14+4\sqrt{10}}}=\frac{\sqrt{6}+3\sqrt{2}+2-\sqrt{6}-2}{2-\sqrt{10}+\sqrt{2}-\sqrt{10}-2}=\frac{3\sqrt{2}}{\sqrt{2}}=3\)
HH
0
NN
28 tháng 10 2020
a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)
\(=\frac{\sqrt{2\left(4-\sqrt{7}\right)}-\sqrt{2\left(4+\sqrt{7}\right)}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+2}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|+2}{\sqrt{2}}=\frac{\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)+2}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1+2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)
b) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)
\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)
\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)
12 tháng 12 2019
a) Đặt \(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(A^2=5-2\sqrt{6}+2\sqrt{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+5+2\sqrt{6}\)
\(=10+2\sqrt{25-4.6}=10+2\sqrt{1}=10+2=12\)
\(\Rightarrow A=\sqrt{12}\)
b)\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\sqrt{2}.\sqrt{2}-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
\(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)
\(A=\sqrt{11-2\sqrt{10}=\sqrt{ }\left(\sqrt{10-1}\right)=\sqrt{ }10-1}\)