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H9
HT.Phong (9A5)
CTVHS
17 tháng 7 2023
1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=2+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}-1\)
2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
PN
3
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 9 2021
\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left|2\sqrt{5}-3\right|-\left|\sqrt{5}-2\right|=2\sqrt{5}-3-\sqrt{5}+2=\sqrt{5}-1\)
12 tháng 9 2021
b)\(=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c)\(=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{5}-3-\sqrt{5}+2\)
\(=\sqrt{5}-1\)
27 tháng 8 2023
a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
=3-4=-1
b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)
c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
NT
3
22 tháng 11 2021
\(\sqrt{5+\sqrt{21}}-\sqrt{5-\sqrt{21}}\\ =\dfrac{\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)}{\sqrt{2}}\\ =\dfrac{\left(\sqrt{7+2\sqrt{7}.\sqrt{3}+3}-\sqrt{7-2\sqrt{7}.\sqrt{3}+3}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 11 2021
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}+\sqrt{3}\right|-\left|\sqrt{7}-\sqrt{3}\right|\right)\)
\(=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
TT
1
HT
3 tháng 9 2017
\(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)
Vậy...
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PN
1
23 tháng 8 2021
\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}=\dfrac{3-\sqrt{5}}{2}\)
5 tháng 7 2021
a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)
\(=\sqrt{\dfrac{1}{2}\left(2+2\sqrt{3}\right)^2}-\sqrt{\dfrac{1}{2}\left(2-2\sqrt{3}\right)^2}\)\(=\sqrt{\dfrac{1}{2}}\left(2+2\sqrt{3}\right)-\sqrt{\dfrac{1}{2}}\left(2\sqrt{3}-2\right)=2\sqrt{2}\)
b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)
5 tháng 7 2021
a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)
\(=\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}\)
\(=2\sqrt{2}\)
b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)
\(=\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)\)
=16-5=11
\(\sqrt{\sqrt{5}-\sqrt{5-\sqrt{21-4\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\sqrt{20^2}-2.\sqrt{20}+1}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\left(\sqrt{20}-1\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\left|\sqrt{20}-1\right|}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{20}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{5^2}-2\sqrt{5}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=1\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)