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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) \(\sqrt{21-6\sqrt{6}}\)

\(=\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+3}\)

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

\(=3\sqrt{2}-\sqrt{3}\)

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) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)

\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)

\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)

\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)

\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)

\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)

10 tháng 6 2016

\(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}-\sqrt{2}=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}-2}{\sqrt{2}}=\frac{\sqrt{11}+1-\left(\sqrt{11}-1\right)-2}{\sqrt{2}}=0\)

15 tháng 6 2017

Đặt \(B=\frac{\sqrt{11+\sqrt{5}}+\sqrt{11-\sqrt{5}}}{\sqrt{11+2\sqrt{29}}}\)Ta có B>0

\(B^2=2\Rightarrow B=\sqrt{2}\)

Vậy \(A=\sqrt{2}+\sqrt{\left(2-\sqrt{2}\right)^2}=2\)

10 tháng 3 2019

\(A=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\) \(\left(\sqrt{6}+11\right)\)

\(A=\left(\frac{3.\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}\right)+\left(\frac{2.\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}{\sqrt{6}-2}\right)-\frac{4.\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\)\(\left(\sqrt{6}+11\right)\)

\(A=\left(3.\left(\sqrt{6}-1\right)+2.\left(\sqrt{6}+2\right)-4.\left(3+\sqrt{6}\right)\right)\left(\sqrt{6}+11\right)\)

\(A=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=-115\)

26 tháng 5 2016

\(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)

26 tháng 5 2016

\(A=\sqrt{11-2\sqrt{10}=\sqrt{ }\left(\sqrt{10-1}\right)=\sqrt{ }10-1}\)