\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
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AH
Akai Haruma
Giáo viên
23 tháng 7 2021
Lời giải:
Đặt biểu thức là $A$. Ta có:
\(A=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{2}.\sqrt{5-\sqrt{21}}\)
\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{10-2\sqrt{21}}\)
\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{(\sqrt{7}-\sqrt{3})^2}\)
\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3})|\sqrt{7}-\sqrt{3}|=(5+\sqrt{21})(\sqrt{7}-\sqrt{3})^2\)
\(=(5+\sqrt{21})(10-2\sqrt{21})=2(5+\sqrt{21})(5-\sqrt{21})=2(5^2-21)=8\)
23 tháng 7 2021
Ta có: \(\left(5+\sqrt{21}\right)\cdot\left(\sqrt{14}-\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)
\(=\dfrac{\left(10+2\sqrt{21}\right)\cdot\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}}{2}\)
\(=\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2\cdot\left(\sqrt{7}-\sqrt{3}\right)^2}{2}\)
=8
4 tháng 8 2016
a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)
20 tháng 6 2021
`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`
`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`
`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`
`=sqrt2(sqrt5+1+sqrt5-1)`
`=sqrt{2}.2sqrt5`
`=2sqrt{10}`
20 tháng 6 2021
`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`
`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`
`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`
`=2(7-3)`
`=8`
`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`
`=2/sqrt2=sqrt2`
17 tháng 6 2021
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
9 tháng 7 2023
b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)
\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)
a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)
=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)
=>\(A=\sqrt{2\sqrt{2}+2}\)
Đặt \(B=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2+\sqrt{2}}\)
=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)
=>\(B^2=2\sqrt{2}+2+2+\sqrt{2}-2\sqrt{\sqrt{2}\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)
=>\(B\simeq0,35\)
4 tháng 10 2017
\(A=\)bn ghi lại đề nha mình lười
\(=\left(\sqrt{5+\sqrt{21}}\right)^2\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{5-\sqrt{21}}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)
\(=\left(\sqrt{\left(5^2-21\right)}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)
\(=2.\left(\sqrt{5+\sqrt{21}}\right)\sqrt{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\left(\sqrt{10+2\sqrt{21}}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\left(\sqrt{7+2\sqrt{21}+3}\right) \left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2.\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=2.\left(7-3\right)=2.4=8\)
tíck mình nha bn thanks nhìu !!!!!!!!!
27 tháng 4 2017
\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{7}+\left(\sqrt{7}\right)^2}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)=2\left(25-21\right)=2\cdot4=8\)
OM
15 tháng 8 2020
\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\sqrt{5+\sqrt{21}}\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{4}\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=2\left(7-3\right)=2.4=8\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
\(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)
\(=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\)
\(=2\left(25-21\right)=8\)
4 tháng 10 2017
\(a)A=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\sqrt{5}+3+\sqrt{3}-\left(\sqrt{5}+3\right)\\ =\sqrt{3}\)
\(b)B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2\\ =\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\\ =2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\\ =2\left(25-21\right)=8\)
\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{7-2\sqrt{7}.\sqrt{3}+3}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)=2\left(25-21\right)=2.4=8\)