Rút Gọn
\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)
K
Khách
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KT
26 tháng 7 2018
\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)
\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)
\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)
\(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)
\(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)
\(\Rightarrow\)\(B=\sqrt{22}\)
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
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}+\sqrt{3}\right|-\left|\sqrt{7}-\sqrt{3}\right|\right)\)
\(=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
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
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 !!!!!!!!!
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) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
N
1
NH
1
DK
9 tháng 8 2016
\(\sqrt{5-\sqrt{21}}+\sqrt{5+\sqrt{21}}\)
\(=\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}\)
\(=\left|\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right|+\left|\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right|\)
\(=\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}+\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\)
\(=2.\sqrt{\frac{7}{2}}\)
\(=\sqrt{14}\)
Chúc bạn học gỏi và tíck cho mìk vs nha!
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: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)
\(=\sqrt{\frac{2\left(5+\sqrt{21}\right)}{2}}+\sqrt{\frac{2\left(5-\sqrt{21}\right)}{2}}\)
\(=\sqrt{\frac{10+2\sqrt{21}}{2}}+\sqrt{\frac{10-2\sqrt{21}}{2}}\)
\(=\sqrt{\frac{3+2\sqrt{21}+7}{2}}+\sqrt{\frac{3-2\sqrt{21+7}}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{3}+\sqrt{7}\right)}{2}}+\sqrt{\frac{\left(\sqrt{3}+\sqrt{7}\right)}{2}}\)
\(=\frac{\left|\sqrt{3}+\sqrt{7}\right|}{2}+\frac{\left|\sqrt{3}-\sqrt{7}\right|}{2}\)
\(=\frac{\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{7}}{2}\)
\(=\sqrt{\frac{6}{2}}=\sqrt{3}\)