\(\sqrt{17-3\sqrt{32}+\sqrt{17+3\sqrt{32}}}\)
làm giúp mình ạ
K
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3 tháng 8 2016
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-2\sqrt{72}}+\sqrt{17+2\sqrt{72}}..\)
= \(\sqrt{9-2\sqrt{9.8}+8}+\sqrt{9+2\sqrt{9.8}+8}.\)
=\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}.\)
= \(\left|3-2\sqrt{2}\right|+3+2\sqrt{2}=3-2\sqrt{2}+3+2\sqrt{2}=6.\)( vì 3 > 2 căn 2 )
TV
2
TH
1
4 tháng 10 2021
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)
TA
6 tháng 8 2021
`\sqrt(17-3\sqrt32)+\sqrt(17+3\sqrt32)`
`=\sqrt(17-12\sqrt2)+\sqrt(17+12\sqrt2)`
`=\sqrt(9-12\sqrt2+8)+\sqrt(9+12\sqrt2+8)`
`=\sqrt(3^2-2.3.2\sqrt2 +(2\sqrt2)^2)+\sqrt(3^2+2.3.2\sqrt2+(2\sqrt2)^2)`
`=\sqrt((3-2\sqrt2)^2)+\sqrt((3+2\sqrt2)^2)`
`=3-2\sqrt2+3+2\sqrt2`
`=6`
6 tháng 8 2021
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
HT
1
HN
8 tháng 10 2021
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{9+2.3.\sqrt{8}+8}+\sqrt{9-2.3.\sqrt{8}+8}\)
\(=\sqrt{\left(3+\sqrt{8}\right)^2}+\sqrt{\left(3-\sqrt{8}\right)^2}=\left|3+\sqrt{8}\right|+\left|3-\sqrt{8}\right|\)
\(=3+\sqrt{8}+3-\sqrt{8}\) (do \(3>\sqrt{8}\))
\(=6\)
AH
Akai Haruma
Giáo viên
29 tháng 7 2021
Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.
** Bài toán rút gọn**
Lời giải:
\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)
\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)
\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)
\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)
\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)
--------------------
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)
\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)
\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)
----------------------
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)
\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)
\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|-\left|4-2\sqrt{2}\right|=3-2\sqrt{2}-4+2\sqrt{2}\)
\(=-1\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|=3-2\sqrt{2}+3+2\sqrt{2}\)
\(=6\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
24 tháng 8 2023
a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)
\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)
\(=2\left(3-2\sqrt{2}\right)=6-4\sqrt{2}\)
4 tháng 7 2017
\(=\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{2}-3\right)^2}-\sqrt{\left(2\sqrt{2}+3\right)^2}\)
\(=-\left(2\sqrt{2}-3\right)-\left(2\sqrt{2}+3\right)=-2\sqrt{2}+3-2\sqrt{2}-3\)
\(=-4\sqrt{2}\)
7 tháng 8 2017
\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}=2\sqrt{17-3\sqrt{32}}\)
\(=\sqrt{4\left(17-3\sqrt{32}\right)}=\sqrt{68-12\sqrt{32}}=\sqrt{36-12\sqrt{32}+32}\)
\(=\sqrt{6^2-2.6.\sqrt{32}+\left(\sqrt{32}\right)^2}=\sqrt{\left(6-\sqrt{32}\right)^2}=\left|6-\sqrt{32}\right|\)
\(=6-\sqrt{32}=6-4\sqrt{2}\)
\(\sqrt{17+3\sqrt{32}}=\sqrt{17+12\sqrt{2}}=\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)\(=\sqrt{\left(3+2\sqrt{2}\right)^2}=3+2\sqrt{2}\)
Biểu thức trở thành \(\sqrt{17-3\sqrt{32}+3+2\sqrt{2}}=\sqrt{20-10\sqrt{2}}\)
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= -2,112.760005