Tính D= |sqrt(8)-3|+|sqrt(19)-4|-(sqrt(19)-sqrt(8))
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HT
14 tháng 6 2019
\(=\left(\sqrt{3}+4\right)\sqrt{\left(4-\sqrt{3}\right)^2}+\left(\sqrt{3}-4\right)\sqrt{\left(4+\sqrt{3}\right)^2}=\left(\sqrt{3}+4\right)\left(4-\sqrt{3}\right)+\left(\sqrt{3}-4\right)\left(4+\sqrt{3}\right)\)
\(=16-3+3-16=0\)
4 tháng 7 2023
a: \(x=4+\sqrt{3}+4-\sqrt{3}=8\)
Khi x=8 thì \(A=\dfrac{2-5\cdot2\sqrt{2}}{2\sqrt{2}+1}=\dfrac{2-10\sqrt{2}}{2\sqrt{2}+1}=-6+2\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
\(A=\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}\)
\(A=\left(4+\sqrt{3}\right)\sqrt{4^2-2\cdot4\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(A=\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}\)
\(A=\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)\)
\(A=4^2-3\)
\(A=13\)
\(B=\dfrac{3}{4+\sqrt{13}}+\dfrac{\sqrt{52}}{2}-3\)
\(B=\dfrac{3\left(4-\sqrt{13}\right)}{\left(4-\sqrt{13}\right)\left(4+\sqrt{13}\right)}+\dfrac{2\sqrt{13}}{2}-3\)
\(B=\dfrac{3\left(4-\sqrt{13}\right)}{16-13}+\sqrt{13}-3\)
\(B=4-\sqrt{13}+\sqrt{13}-3\)
\(B=4-3\)
\(B=1\)
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
\(A=\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5}^2-2\sqrt{5}+1^2}=\sqrt{ \left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)
\(D=\sqrt{5-2\sqrt{6}}=\sqrt{5-2.\sqrt{2}.\sqrt{3}}=\sqrt{\sqrt{3}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{2^2}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
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.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
TB
2
NC
29 tháng 8 2020
a) \(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(A=\sqrt{16+8\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}\)
\(A=\sqrt{\left(4+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(A=4+\sqrt{3}-\sqrt{3}-1=3\)
b) \(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(B=\sqrt{25+10\sqrt{2}+2}-\sqrt{16+8\sqrt{2}+2}\)
\(A=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(A=5+\sqrt{2}-4-\sqrt{2}=1\)
29 tháng 8 2020
\(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3+8\sqrt{3}+16}-\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot4+4^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}+1^2}\)
\(=\sqrt{\left(\sqrt{3}+4\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}+4\right|-\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}+4-\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}+4-\sqrt{3}-1=3\)
\(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=\sqrt{2+10\sqrt{2}+25}-\sqrt{2+8\sqrt{2}+16}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot5+5^2}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot4+4^2}\)
\(=\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{\left(\sqrt{2}+4\right)^2}\)
\(=\left|\sqrt{2}+5\right|-\left|\sqrt{2}+4\right|\)
\(=\sqrt{2}+5-\left(\sqrt{2}+4\right)\)
\(=\sqrt{2}+5-\sqrt{2}-4=1\)
BT
2 tháng 8 2017
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
11 tháng 8 2019
b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))
t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).\(\sqrt{2}\)(\(\sqrt{5}\) -1)
t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))
t = (\(\sqrt{5}\) -1)2.(3 +\(\sqrt{5}\))
t = (5 - \(2\sqrt{5}\)+1).(3 +\(\sqrt{5}\))
t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)
t = 6
Sqrt ở đây làm căn bậc hai nhé
\(D=\left|\sqrt{8}-3\right|+\left|\sqrt{19}-4\right|-\left(\sqrt{19}-\sqrt{8}\right)\)
\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-\sqrt{8}\right)\)
\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)
\(D=-2\sqrt{2}+3+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)
\(D=-2\sqrt{2}+3+\sqrt{19}-4-\sqrt{19}+2\sqrt{2}\)
\(D=-2\sqrt{2}+3-4+2\sqrt{2}\)
\(D=3-4\)
\(D=-1\)