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bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon
a) \(2^{2017}+2^{2014}=2^{2014}\left(2^3+1\right)=2^{2014}.9⋮9\)
b) \(4^{2016}+4^{2014}=4^{2014}\left(4^2+1\right)=4^{2014}.17\)
2) \(3.4^{n+2}+4^n=49\\ \Rightarrow4^n\left(3.4^2+1\right)=49\\ \Rightarrow4^n.33=49\\ \Rightarrow4^n=16\\ \Rightarrow n=2\)
3) \(200-180:\left[36.5-7.25\right]\\ =200-180:\left[180-175\right]\\ =200-180:5\\ =200-36\\ =164\)
1/
a. \(x^3-2=25\)
\(x^3=25+2\)
\(x^3=27\)
\(\Rightarrow x=3\)
b.\(\left(x-3\right)^2=25\)
\(\left(x-3\right)^2=5^2\)
\(\Rightarrow x-3=5\)
\(\Rightarrow x=8\)
1,a, x^3-2=25 b, (x-3)^2=25 c, x^3-x^2=55 d,[(8.x-12):4].3^7=3^10
x^3=27 (x-3)^2=5^2 không có giá trị x (8.x-12):4=3^3
x^3=3^3 x-3=5 8.x-12=108
x=3 x=8 8.x=120
x=15
2, a, \(7^6:7^4+3^4.3^2-3^7:3\) b, 1736-(21-16).32+6.7^2 c,56.17+17.44-4^3.5+6.(3^2-2)
=\(7^2+3^6-3^6\) =1736-5.32+6.49 =17.(56+44)-320+42
=\(49\) =1736-160+294 =17.10-278
=1736+134 =170-278
=1870 =-108
d, 3.10^2-[1200-(4^2-2.3)^3]
=300-[1200-(16-6)^3]
=300-(1200-10^3)
=300-(1200-1000)
=300-200
=100
B=1+4+4^2+4^3+......+4^100
4B=4+4^2+4^3+4^4+........+4^101
4B - B = 4^101-1
3B=4^101-1
B=(4^101-1):3
a: =18x941+18x59
=18(941+59)
=18x1000=18000
b: \(=81:27-16:8=3-2=1\)
c: =30-40+25=-10+25=15
d: =17(85+15)-150=1700-150=1550
e: =-150-180-200=-530
f: =17+15+40=72
a) \(A=1+3+3^2+.....+3^{10}⋮4\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+.......+\left(3^9+3^{10}\right)\)
\(=\left(1+3\right)+\left(3^2\cdot1+3^2\cdot3\right)+.....+\left(3^9\cdot1+3^9\cdot3\right)\)
\(=\left(1+3\right)+3^2\left(1+3\right)+....+3^9\left(1+3\right)\)
\(=4\cdot1+3^2\cdot4+.......+3^9\cdot4\)
\(=4\cdot\left(1+3^2+.....+3^9\right)⋮4\)
Do đó A \(⋮\) 4
b) \(B=16^5+2^{15}⋮33\)
Ta có \(B=16^5+2^{15}\)
\(=\left(2^4\right)^5+2^{15}\)
\(=2^{20}+2^{15}\)
\(=2^{15}\cdot2^5+2^{15}\cdot1\)
\(=2^{15}\cdot\left(2^5+1\right)\)
\(=2^5\cdot\left(32+1\right)\)
\(=2^{15}\cdot33⋮33\)
Do đó \(B⋮33\)
`a)` Ta có:
`C=1+4+4^2+4^3+4^4+4^5+4^6`
`->4C=4(1+4+4^2+4^3+4^4+4^5+4^6)`
`=4*1+4*4+4*4^2+4*4^3+4*4^4+4*4^5+4*4^6`
`=4+4^2+4^3+4^4+4^5+4^6+4^7`
Vậy: `4C=4+4^2+4^3+4^4+4^5+4^6+4^7`
`b)` Theo câu `a):`
`4C=4+4^2+4^3+4^4+4^5+4^6+4^7`
`->4C-C=(4+4^2+4^3+4^4+4^5+4^6+4^7)-(1+4+4^2+4^3+4^4+4^5+4^6)`
`->3C=4+4^2+4^3+4^4+4^5+4^6+4^7-1-4-4^2-4^3-4^4-4^5-4^6`
`->3C=4^7-1`
`->C=(4^7-1)/3` (đpcm)
a) Ta có:
\(C=1+4+4^2+4^3+4^4+4^5+4^6\)
\(4C=\left(1+4+4^2+4^3+4^4+4^5+4^6\right)\cdot4\)
\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)
b)
\(\Rightarrow3C=4C-C=\left(4+4^2+4^3+4^4+4^5+4^6+4^7\right)-\left(1+4+4^2+4^3+4^4+4^5+4^6\right)\)
\(3C=4^7-1\)
Vì \(3C\vdots3\) nên \(4^7-1\vdots3\)
Vậy ....