Rút gọn: \(\sqrt{21-8\sqrt{5}}-\sqrt{21-4\sqrt{5}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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
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) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
TN
2
Y
21 tháng 6 2023
\(\sqrt{\sqrt{5}-\sqrt{5-\sqrt{21-4\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\sqrt{20^2}-2.\sqrt{20}+1}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\left(\sqrt{20}-1\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\left|\sqrt{20}-1\right|}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{20}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{5^2}-2\sqrt{5}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=1\)
21 tháng 6 2023
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
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
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}+\sqrt{3}\right|-\left|\sqrt{7}-\sqrt{3}\right|\right)\)
\(=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
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
21 tháng 8 2018
\(\sqrt{6+2\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{16-2\cdot4\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}\)
\(=\sqrt{5}+1+4-\sqrt{5}\)( vì \(\sqrt{5}+1>0\)và \(4-\sqrt{5}>0\))
\(=5\)
IS
21 tháng 8 2018
\(\sqrt{6+2\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{16+8\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}\)
\(=\left|\sqrt{5}+1\right|+\left|4-\sqrt{5}\right|\)
\(=\sqrt{5}+1+4-\sqrt{5}\)
\(=5\)
N
1
KT
26 tháng 7 2018
\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)
\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)
\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)
\(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)
\(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)
\(\Rightarrow\)\(B=\sqrt{22}\)
\(\sqrt{21-8\sqrt{5}}\)\(-\sqrt{21-4\sqrt{5}}\)
\(=\sqrt{16-2.4\sqrt{5}+5}\)\(-\sqrt{20-2\sqrt{20}+1}\)
\(=\sqrt{\left(4-\sqrt{5}\right)^2}\)\(-\sqrt{\left(\sqrt{20}-1\right)}\)
\(=4-\sqrt{5}-\left(\sqrt{20}-1\right)\)
\(=4-\sqrt{5}-\sqrt{20}+1\)
\(=5-\sqrt{5}-2\sqrt{5}\)
\(=5-3\sqrt{5}\)