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HL
1
9 tháng 12 2015
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1-2\sqrt{2}+1=\sqrt{2}\)
MN
7 tháng 8 2020
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
29 tháng 7 2017
a. \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}=\sqrt{13+6\sqrt{4+\sqrt{\left(\sqrt{8}-1\right)^2}}}=\sqrt{13+6\sqrt{4+\sqrt{8}-1}}=\sqrt{13+6\sqrt{3+\sqrt{8}}}=\sqrt{13+6\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+6\left(\sqrt{2}+1\right)}=\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}+1\right)^2}=1+3\sqrt{2}\)
b. \(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}}-4}=\left(\sqrt{3}-1\right)\sqrt{2\sqrt{\left(4+\sqrt{3}\right)^2}-4}=\left(\sqrt{3}-1\right)\sqrt{8+2\sqrt{3}-4}=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
c. \(\sqrt{5+2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}\)
d. \(\sqrt{5-2\sqrt{6}}+\sqrt{11-4\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}=\sqrt{3}-\sqrt{2}+2\sqrt{2}-\sqrt{3}=\sqrt{2}\)
20 tháng 7 2016
\(\frac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}\)
= \(\frac{\sqrt{2^2-2\sqrt{5}2+\sqrt{5^2}}}{2-\sqrt{5}}\)
= \(\frac{\sqrt{\left(2-\sqrt{5}\right)^2}}{2-\sqrt{5}}\)
= \(\frac{\sqrt{5}-2}{2-\sqrt{5}}\)
= -1
Chúc bạn làm bài tốt :)
6 tháng 8 2020
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Học tốt
6 tháng 8 2020
Bài làm:
a) \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(B=2+\sqrt{5}-\sqrt{5}+2\)
\(B=4\)
6 tháng 8 2017
a, Đặt biểu thức là A
<=>\(\sqrt{2}\)A = \(\sqrt{4+2\sqrt{3}}\)- 2 . \(\sqrt{3}\)+1
= \(\sqrt{\left(\sqrt{3}+1\right)^2}\) - 2.\(\sqrt{3}\)+1 = \(\sqrt{3}\)+ 1 - \(2\sqrt{3}\) + 1 = 2-\(\sqrt{3}\)
B
12 tháng 7 2019
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
12 tháng 7 2019
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
\(\sqrt{6-4\sqrt{2}}-\sqrt{19-6\sqrt{2}}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{18}-1\right)^2}\\ =2-\sqrt{2}-\sqrt{18}+1\\ =3-4\sqrt{2}\)