\(\left(\sqrt{4^2}+\sqrt{\left(-4\right)^2}\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}.\)
K
Khách
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26 tháng 6 2021
\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)
\(=54+8-32=30\)
\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)
\(=5-2\sqrt{2}\)
\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)
\(=2-2\sqrt{3}\)
\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)
\(=2\sqrt{6}\)
\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)
26 tháng 6 2021
`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2019
1/ \(=2+\sqrt{5}-\left|2-\sqrt{5}\right|=2+\sqrt{5}-\sqrt{5}+2=4\)
2/ bạn coi lại đề
3/ \(=\sqrt{2}+1-\left|1-\sqrt{2}\right|=\sqrt{2}+1-\sqrt{2}+1=2\)
4/ \(=\sqrt{3}+2-\left|\sqrt{3}-2\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
5/ \(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
6/ \(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
13 tháng 6 2019
Các bạn giúp mình với, tối nay mình nộp rồi.
Câu 6 sửa lại đề giúp mình như này nhé:
\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
29 tháng 6 2023
b: \(=16-2\cdot4\cdot2\sqrt{5}+20-9-4\sqrt{5}\)
=27-20căn 5
a: 2-4căn 3<0
nên biểu thức ko có giá trị
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 6 2023
\(b,\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\\ =\left[\left(4-2\sqrt{5}\right)-\left(\sqrt{5}+2\right)\right].\left[\left(4-2\sqrt{5}\right)+\left(\sqrt{5}+2\right)\right]=\left(2-3\sqrt{5}\right)\left(6-\sqrt{5}\right)\)
8 tháng 6 2021
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
---------------------
$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
H
30 tháng 9 2023
\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
__
C làm giống câu a, nhé.
__
\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
__
\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
15 tháng 5 2021
`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`
`=`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`
`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`
`=2(\sqrt5-1)sqrt{6+2\sqrt5}`
`=2(\sqrt5-1)(\sqrt5+1)`
`=2(5-1)`
`=8`
15 tháng 5 2021
`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`
`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`
`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`
`=2(\sqrt5-1)sqrt{6+2\sqrt5}`
`=2(\sqrt5-1)(\sqrt5+1)`
`=2(5-1)`
`=8`
`(4\sqrt2+\sqrt{30})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`
`=\sqrt2(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`
`=(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{8-2\sqrt{15}}`
`=(4+\sqrt{15})(\sqrt5-\sqrt3)(\sqrt5-\sqrt3)`
`=(4+\sqrt{15})(8-2\sqrt{15})`
`=2(4+\sqrt{15})(4-\sqrt{15})`
`=2(16-15)`
`=2`
4+4.6-9=8.-3=-24
\(\left(\sqrt{4^2}+\sqrt{\left(-4\right)^2}\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}\)
\(=\left(4+4\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}\)
\(=8.\sqrt{4^{-3}}-\sqrt{3^{-4}}\)
\(=8.\sqrt{\frac{1}{4^3}}-\sqrt{\frac{1}{3^4}}\)
\(=8.\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)