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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) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)
\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)
\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)
23 tháng 9 2017
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
22 tháng 7 2016
a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
B
12 tháng 7 2019
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
12 tháng 7 2019
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
1 tháng 10 2019
\(A=\frac{1}{\sqrt{11-2\sqrt{30}}}-\frac{3}{\sqrt{7-2\sqrt{10}}}+\frac{4}{\sqrt{8+4\sqrt{3}}}\)
\(=\frac{1}{\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}-\frac{3}{\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}}+\frac{2}{\sqrt{4+2\sqrt{3}}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}-\frac{3}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{2}{\sqrt{3}+1}\)
\(=\frac{6-5}{\sqrt{6}-\sqrt{5}}-\frac{5-2}{\sqrt{5}-\sqrt{2}}+\frac{3-1}{\sqrt{3}+1}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}{\sqrt{6}-\sqrt{5}}-\frac{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}+\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{3}+1}\)
\(=\sqrt{6}+\sqrt{5}-\sqrt{5}+\sqrt{2}+\sqrt{3}+1=\sqrt{6}+\sqrt{2}+\sqrt{3}+1\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)+\sqrt{3}+1=\left(\sqrt{3}+1\right)\left(\sqrt{2}+1\right)\)
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=\(\frac{\sqrt{6}+3\sqrt{2}+2-\sqrt{10+4\sqrt{6}}}{2+\sqrt{10}+\sqrt{2}-\sqrt{14+4\sqrt{10}}}=\frac{\sqrt{6}+3\sqrt{2}+2-\sqrt{6}-2}{2-\sqrt{10}+\sqrt{2}-\sqrt{10}-2}=\frac{3\sqrt{2}}{\sqrt{2}}=3\)
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)\(\Rightarrow\sqrt{2}\left[\left(\sqrt{4-\sqrt{7}}\right)-\left(\sqrt{4+\sqrt{7}}\right)\right]=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}=\sqrt{7}-1-1-\sqrt{7}=-2\Rightarrow\frac{-2}{\sqrt{2}}=-\sqrt{2}\)