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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 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
T
2
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
31 tháng 8 2021
\(\left(\sqrt{14}+\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
=7-3
=4
25 tháng 9 2021
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
DV
1
17 tháng 7 2018
\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{7-2\sqrt{7}.\sqrt{3}+3}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)=2\left(25-21\right)=2.4=8\)
PT
1
27 tháng 4 2017
\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{7}+\left(\sqrt{7}\right)^2}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)=2\left(25-21\right)=2\cdot4=8\)
OS
2
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
\(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)
\(=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\)
\(=2\left(25-21\right)=8\)
31 tháng 8 2017
1) \(5-2\sqrt{6}=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
2) \(8+2\sqrt{15}=\left(\sqrt{5}\right)^2+2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{5}+\sqrt{3}\right)^2\)
3) \(10-2\sqrt{21}=\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{7}-\sqrt{3}\right)^2\)
4) \(21+6\sqrt{6}=\left(\sqrt{18}\right)^2+2.\sqrt{18}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{18}+\sqrt{3}\right)^2\)
5) \(14+8\sqrt{3}=\left(\sqrt{8}\right)^2+2.\sqrt{8}.\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{8}+\sqrt{6}\right)^2\)
6) \(36-12\sqrt{5}=\left(\sqrt{30}\right)^2-2.\sqrt{30}.\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{30}-\sqrt{6}\right)^2\)
7) \(25+4\sqrt{6}=\left(\sqrt{24}\right)^2+2\sqrt{24}.1+1^2=\left(\sqrt{24}+1\right)^2\)
8) \(98-16\sqrt{3}=\left(\sqrt{96}\right)^2-2\sqrt{96}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{96}-\sqrt{2}\right)^2\)
BT
2 tháng 8 2017
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
\(\sqrt{14}-6\sqrt{5}-\sqrt{21}-8\sqrt{5}.\)
\(=\left(-6\sqrt{5}-8\sqrt{5}\right)+\left(\sqrt{14}-\sqrt{21}\right)\)
\(=-14\sqrt{5}+\sqrt{7}\left(\sqrt{2}-\sqrt{3}\right)\)