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TV
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
\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}=2\sqrt{17-3\sqrt{32}}\)
\(=\sqrt{4\left(17-3\sqrt{32}\right)}=\sqrt{68-12\sqrt{32}}=\sqrt{36-12\sqrt{32}+32}\)
\(=\sqrt{6^2-2.6.\sqrt{32}+\left(\sqrt{32}\right)^2}=\sqrt{\left(6-\sqrt{32}\right)^2}=\left|6-\sqrt{32}\right|\)
\(=6-\sqrt{32}=6-4\sqrt{2}\)
9 tháng 8 2016
a) \(\sqrt{28}+\sqrt{125}-3\sqrt{343}-\frac{3}{8}\sqrt{396}=2\sqrt{7}+5\sqrt{5}-21\sqrt{7}-\frac{9\sqrt{11}}{4}\)
\(=-19\sqrt{7}+5\sqrt{5}-\frac{9\sqrt{11}}{4}\)
b) \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\left(3-2\sqrt{2}\right)+2\sqrt{2}+3=6\)
MT
24 tháng 7 2015
\(\sqrt{17-3\sqrt{32}}-\sqrt{17+3\sqrt{32}}=\frac{\sqrt{2}\sqrt{17-3\sqrt{32}}-\sqrt{2}\sqrt{17+3\sqrt{32}}}{\sqrt{2}}\)
\(=\frac{\sqrt{34-2.3\sqrt{32}}-\sqrt{34+2.3\sqrt{32}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}.4+4^2}-\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.4+4^2}}{\sqrt{2}}\)
\(=\frac{3\sqrt{2}-4-3\sqrt{2}-4}{\sqrt{2}}=\frac{-8}{\sqrt{2}}=-4\sqrt{2}\)
N
3
17 tháng 10 2017
\(\sqrt{17+3\sqrt{32}}=\sqrt{17+12\sqrt{2}}=\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)\(=\sqrt{\left(3+2\sqrt{2}\right)^2}=3+2\sqrt{2}\)
Biểu thức trở thành \(\sqrt{17-3\sqrt{32}+3+2\sqrt{2}}=\sqrt{20-10\sqrt{2}}\)
Bạn xem lại đề rồi đăng lại
N
22 tháng 6 2018
Bài làm của: Phùng Khánh Linh
c)\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
= \(\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{4^2-2.4.\sqrt{8}+\left(\sqrt{8}\right)^2}\)
= \(\sqrt{\left(3-2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{\left(4-\sqrt{8}\right)^2}\)
= \(\left|3-2\sqrt{2}\right|-\left|4-\sqrt{8}\right|\)
= (3 - 2\(\sqrt{2}\)) - (4 - \(\sqrt{8}\))
= 3 - 2\(\sqrt{2}\) - 4 + \(\sqrt{8}\)
= -1
22 tháng 6 2018
\(a.\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\text{|}\sqrt{3}+1\text{|}-\text{|}\sqrt{3}-1\text{|}=2\)\(b.\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\text{|}\sqrt{5}-2\text{|}-\text{|}\sqrt{5}+2\text{|}=-4\) Còn lại tương tự nhé .
M
14 tháng 8 2020
a, \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
= \(\sqrt{3^2-2.3.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{6^2-2.6.\sqrt{6}+\left(\sqrt{6}\right)^2}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(6-\sqrt{6}\right)^2}\)
= \(\left|3-\sqrt{6}\right|+\left|6-\sqrt{6}\right|\)
= \(3-\sqrt{6}+6-\sqrt{6}\)
= \(9-2\sqrt{6}\)
b. Đặt B = \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
Nhận xét : B > 0 , bình phương hai vế ta được :
\(B^2=\left(\sqrt{17-3\sqrt{32}}\right)^2+\left(\sqrt{17+3\sqrt{32}}\right)^2\)
\(B^2=17-3\sqrt{32}+17+3\sqrt{32}+2\sqrt{\left(17-3\sqrt{32}\right)\left(17+3\sqrt{32}\right)}\)
\(B^2=34+2\sqrt{17^2-\left(3\sqrt{32}\right)^2}\)
\(B^2=34+2\sqrt{289-288}\)
\(B^2=34+2=36\)
=> \(B=\pm\sqrt{36}\) mà B > 0 nên \(B=\sqrt{36}=6\)
c, Đặt C = \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
Nhận xét : C > 0 , bình phương hai vế ta đươc :
\(C^2=\left(\sqrt{49-5\sqrt{96}}\right)^2+\left(\sqrt{49+5\sqrt{96}}\right)^2\)
\(C^2=49-5\sqrt{96}+49+5\sqrt{96}+2\sqrt{\left(49-5\sqrt{96}\right)\left(49+5\sqrt{96}\right)}\)
\(C^2=98+2\sqrt{49^2-\left(5\sqrt{96}\right)^2}\)
\(C^2=98+2\sqrt{2401-2400}\)
\(C^2=98+2=100\)
=> \(C=\pm\sqrt{100}\) mà C > 0 nên \(C=\sqrt{100}=10\)
14 tháng 8 2020
a) Ta có: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot3\sqrt{3}\cdot2\sqrt{2}+8}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|3\sqrt{3}-2\sqrt{2}\right|\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)(Vì \(\left\{{}\begin{matrix}3>\sqrt{6}\\3\sqrt{3}>2\sqrt{2}\end{matrix}\right.\))
b) Ta có: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\frac{\sqrt{34-6\sqrt{32}}+\sqrt{34+6\sqrt{32}}}{\sqrt{2}}\)
\(=\frac{\sqrt{18-2\cdot3\sqrt{2}\cdot4+16}+\sqrt{18+2\cdot3\sqrt{2}\cdot4+16}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{2}-4\right)^2}+\sqrt{\left(3\sqrt{2}+4\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|3\sqrt{2}-4\right|+\left|3\sqrt{2}+4\right|}{\sqrt{2}}\)
\(=\frac{3\sqrt{2}-4+3\sqrt{2}+4}{\sqrt{2}}\)(Vì \(3\sqrt{2}>4>0\))
\(=\frac{6\sqrt{2}}{\sqrt{2}}=6\)
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 )
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{9+2.3.\sqrt{8}+8}+\sqrt{9-2.3.\sqrt{8}+8}\)
\(=\sqrt{\left(3+\sqrt{8}\right)^2}+\sqrt{\left(3-\sqrt{8}\right)^2}=\left|3+\sqrt{8}\right|+\left|3-\sqrt{8}\right|\)
\(=3+\sqrt{8}+3-\sqrt{8}\) (do \(3>\sqrt{8}\))
\(=6\)