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6 tháng 7 2017
a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\)
b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)
c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
3 tháng 7 2019
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\sqrt{2}+1\)
G
3 tháng 7 2019
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
NT
1
1 tháng 12 2016
\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}=\sqrt{8\left(3+2\sqrt{2}\right)}-\sqrt{8\left(3-2\sqrt{2}\right)}\)
\(=\sqrt{8}.\left[\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\right]=\sqrt{8}.\left(\sqrt{2}+1-\sqrt{2}+1\right)=2\sqrt{8}=4\sqrt{2}\)
LN
2
11 tháng 6 2018
\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
\(=\sqrt{\left(4+2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=4+2\sqrt{2}-4+2\sqrt{2}\)
\(=4\sqrt{2}\)
HH
11 tháng 6 2018
Giải:
\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
\(=\sqrt{8+2.4.2\sqrt{2}+16}-\sqrt{16-2.4.2\sqrt{2}+8}\)
\(=\sqrt{\left(2\sqrt{2}+4\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}+4-\left(4-2\sqrt{2}\right)\)
\(=2\sqrt{2}+4-4+2\sqrt{2}\)
\(=4\sqrt{2}\)
Vậy ...
HN
0
HR
30 tháng 11 2017
\(8\sqrt{2}\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)\)
\(=8\sqrt{2}\left(\sqrt{16+2.4.\sqrt{8}+8}-\sqrt{16-2.4\sqrt{8}+8}\right)\)
\(=8\sqrt{2}\left(\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}\right)\)
\(=8\sqrt{2}\left(4+\sqrt{8}-4+\sqrt{8}\right)\)
\(=8\sqrt{2}.2\sqrt{8}\)
= 64
DT
11 tháng 12 2017
ta có\(8\sqrt{2}\cdot\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)=8\sqrt{2}\cdot\left(\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}\right)=8\sqrt{2}\cdot\left(4+\sqrt{8}-4+\sqrt{8}\right)=8\sqrt{2}\cdot2\sqrt{8}=64\)vây..................
17 tháng 7 2017
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
14 tháng 7 2017
A=\(\sqrt{\left(4+\sqrt{8}\right)^2}\)\(-\sqrt{\left(4-\sqrt{8}\right)^2}\)=\(4+\sqrt{8}\)\(-\left(4-\sqrt{8}\right)\)=\(2\sqrt{8}\)
Giờ mình chỉ giải đc câu a thôi để hồi nao mình rảnh giải típ cho
Đặt \(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
\(A^2=\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)^2\)
\(A^2=\left(\sqrt{24+16\sqrt{2}}\right)^2-2\sqrt{24+16\sqrt{2}}\cdot\sqrt{24-16\sqrt{2}}+\left(\sqrt{24-16\sqrt{2}}\right)^2\)
\(A^2=\left|24+16\sqrt{2}\right|-2\sqrt{\left(24+16\sqrt{2}\right)\left(24-16\sqrt{2}\right)}+\left|24-16\sqrt{2}\right|\)
\(A^2=24+16\sqrt{2}-2\sqrt{24^2-\left(16\sqrt{2}\right)^2}+24-16\sqrt{2}\)
\(A^2=48-2\sqrt{576-512}\)
\(A^2=48-2\sqrt{64}\)
\(A^2=48-2\sqrt{8^2}=48-2\cdot\left|8\right|=32\)
=> \(A=\sqrt{32}\)