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PN
1
26 tháng 6 2019
\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=2\sqrt{3}-1+2\sqrt{3}+1=4\sqrt{3}\)
\(\sqrt{19-6\sqrt{2}}-\sqrt{19+6\sqrt{2}}=3\sqrt{2}-1-3\sqrt{2}-1=-2\)
11 tháng 7 2021
a) \(\sqrt{19-6\sqrt{2}}=3\sqrt{2}-1\)
b) \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)
d) \(\sqrt{21+12\sqrt{3}}=2\sqrt{3}+3\)
e) \(\sqrt{57-40\sqrt{2}}=4\sqrt{2}-5\)
2
23 tháng 5 2023
`A=\sqrt{6-2\sqrt{5}}`
`A=\sqrt{(\sqrt{5}-1)^2}`
`A=\sqrt{5}-1`
_________
`B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}`
`B=\sqrt{(\sqrt{3}-1)^2}`
`B=\sqrt{3}-1`
_________
`C=\sqrt{19-8\sqrt{3}}`
`C=\sqrt{(4-\sqrt{3})^2}`
`C=4-\sqrt{3}`
_________
`D=\sqrt{5-2\sqrt{6}}`
`D=\sqrt{(\sqrt{3}-\sqrt{2})^2}`
`D=\sqrt{3}-\sqrt{2}`
Y
23 tháng 5 2023
\(A=\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5}^2-2\sqrt{5}+1^2}=\sqrt{ \left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)
\(D=\sqrt{5-2\sqrt{6}}=\sqrt{5-2.\sqrt{2}.\sqrt{3}}=\sqrt{\sqrt{3}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{2^2}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
24 tháng 9 2021
\(a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3\)
24 tháng 9 2021
a) \(=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=-1+\sqrt{6}\)
b) \(=\left|3-2\sqrt{2}\right|+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\)
c) \(=\sqrt{\left(\sqrt{5}+2\right)^2}-\left|1-\sqrt{5}\right|=\sqrt{5}+2+1-\sqrt{5}=3\)
12 tháng 7 2021
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
MH
1
7 tháng 7 2021
\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)
\(=\left|4-3\sqrt{2}\right|-\sqrt{\left(3\sqrt{2}+1\right)^2}\)
\(=3\sqrt{2}-4-\left(3\sqrt{2}+1\right)\)
\(=-5\)
1 tháng 10 2021
a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
\(=5-\sqrt{19}-\sqrt{19}+4\)
\(=9-2\sqrt{19}\)
b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
\(=3-2\sqrt{2}-3+2\sqrt{2}\)
=0
AH
Akai Haruma
Giáo viên
2 tháng 10 2021
c.
Căn bậc 2 không xác định do $2-\sqrt{5}< 0$
d.
\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)
e.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
H
30 tháng 9 2023
\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)
__
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)
__
\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)
AT
12 tháng 7 2021
\(\sqrt{29-4\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}=\sqrt{\left(2\sqrt{7}-1\right)^2}=\left|2\sqrt{7}-1\right|\)
\(=2\sqrt{7}-1\)
\(\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.1+1^2}=\sqrt{\left(3\sqrt{2}+1\right)^2}=\left|3\sqrt{2}+1\right|\)
\(=3\sqrt{2}+1\)
\(\sqrt{28-6\sqrt{3}}=\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.1+1^2}=\sqrt{\left(3\sqrt{3}-1\right)^2}=\left|3\sqrt{3}-1\right|\)
\(=3\sqrt{3}-1\)
\(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}\right)^2-2.3\sqrt{5}.1+1^2}=\sqrt{\left(3\sqrt{5}-1\right)^2}=\left|3\sqrt{5}-1\right|\)
\(=3\sqrt{5}-1\)
\(\sqrt{49+8\sqrt{3}}=\sqrt{\left(4\sqrt{3}\right)^2+2.4\sqrt{3}.1+1^2}=\sqrt{\left(4\sqrt{3}+1\right)^2}=\left|4\sqrt{3}+1\right|\)
\(=4\sqrt{3}+1\)
\(\sqrt{32-8\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.2+2^2}=\sqrt{\left(2\sqrt{7}-2\right)^2}=\left|2\sqrt{7}-2\right|\)
\(=2\sqrt{7}-2\)
12 tháng 7 2021
\(\sqrt{29-4\sqrt{7}}=2\sqrt{7}-1\)
\(\sqrt{19+6\sqrt{2}}=3\sqrt{2}+1\)
\(\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)
\(\sqrt{46-6\sqrt{5}}=3\sqrt{5}-1\)
\(\sqrt{49+8\sqrt{3}}=4\sqrt{3}+1\)
\(\sqrt{32-8\sqrt{7}}=2\sqrt{7}-2\)
LT
1
\(\sqrt{19-6\sqrt{2}}\)
\(=\sqrt{18-2.3\sqrt{2}+1}\)
\(=\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}+1}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1\)