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NH
1
TN
17 tháng 6 2019
\(a,\sqrt{\sqrt{17+12\sqrt{2}}}\)
\(=\sqrt{\sqrt{8+12\sqrt{2}+9}}\)
\(=\sqrt{\sqrt{\left[2\sqrt{2}+3\right]^2}}\)
\(=\sqrt{2\sqrt{2}+3}\)
\(=\sqrt{1+2\sqrt{2}+2}\)
\(=\sqrt{\left[1+\sqrt{2}\right]^2}\)
\(=1+\sqrt{2}\)
\(b,\sqrt{4+2\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{12-12\sqrt{3}+9}\)
\(=\sqrt{\left[1+\sqrt{3}\right]^2}-\sqrt{\left[2\sqrt{3}-3\right]^2}\)
\(=\left(1+\sqrt{3}\right)-\left(2\sqrt{3}-3\right)\)
\(=1+\sqrt{3}-2\sqrt{3}+3\)
\(=4-\sqrt{3}\)
chúc bn học tốt
21 tháng 7 2023
`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.
`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.
`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.
`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.
`= 7`.
`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`
`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`
`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`
`= 33 - 11 = 22`.
N
1
TH
Thầy Hùng Olm
Manager
VIP
5 tháng 7 2023
a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)
4 tháng 9 2016
1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
4 tháng 9 2016
1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)
\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2022
\(A=\sqrt{\left(9\sqrt{2}+2\sqrt{3}\right)^2}-\sqrt{\left(9\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\left|9\sqrt{2}+2\sqrt{3}\right|-\left|9\sqrt{2}-\sqrt{3}\right|\)
\(=9\sqrt{2}+2\sqrt{3}-9\sqrt{2}+\sqrt{3}=3\sqrt{3}\)
Kiểm tra lại đề bài câu B, chỗ \(\sqrt{2+\sqrt{2+2}}\)
24 tháng 9 2021
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
14 tháng 8 2017
a) \(\sqrt{39-12\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{36-12\sqrt{3}+3}+\sqrt{9-12\sqrt{3}+12}\)
\(=\sqrt{\left(6-\sqrt{3}\right)^2}+\sqrt{\left(3-\sqrt{12}\right)^2}\)
\(=6-\sqrt{3}+\sqrt{12}-3=3+\sqrt{3}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
8 tháng 7 2018
\(\sqrt{3+\sqrt{5}}=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}+1}{\sqrt{2}}\)
\(\sqrt{7+3\sqrt{5}}=\frac{\sqrt{14+2.3\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{9+2.3\sqrt{5}+5}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}=\frac{3+\sqrt{5}}{\sqrt{2}}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{3+2.\sqrt{3}.3\sqrt{2}+18}=\sqrt{\left(\sqrt{3}+3\sqrt{2}\right)^2}=\sqrt{3}+3\sqrt{2}\)
\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=3\sqrt{2}-\sqrt{3}\)
Nên \(E=\frac{\sqrt{5}+1+3+\sqrt{5}}{\sqrt{2}}.\left(3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\right)\)
\(=\frac{4+2\sqrt{5}}{\sqrt{2}}.2.3.\sqrt{2}=24+12\sqrt{5}\)
\(\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{12-2.2\sqrt{3}.3+9}\)
\(=\sqrt{\left(\sqrt{12}-3\right)^2}\)
\(=2\sqrt{3}-3\)
\(\sqrt{21-12\sqrt{3}}=\sqrt{\left(2\sqrt{3}-3\right)^2}=\left|2\sqrt{3}-3\right|=2\sqrt{3}-3\)