Cho A=73+74+75+76+.........+797+798. Chứng tỏ A chia hết cho 8.
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\(A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8\)
\(A=\left(7+7^3\right)+\left(7^2+7^4\right)+\left(7^5+7^7\right)+\left(7^6+7^8\right)\)
\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)
\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)
\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)
\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)
Ta có: 5 ⋮ 5
⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm)
A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78
A = (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)
A = 7.(1 + 72) + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)
A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)
A = 7.50 + 72.50 + 75.50 + 76.50
A = 50.(7 + 72 + 75 + 76)
Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm
\(a,=7^4\left(7^2+7-1\right)=7^4\cdot55=7^4\cdot5\cdot11⋮11\)
a) \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4\left(49+7-1\right)=7^4.55⋮55\)
b) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\left(32+1\right)=2^{15}.33⋮33\)
c) \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5=3^{22}.3^4.5=3^{22}.405⋮405\)
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33⋮33\)
c: \(=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}\cdot5=3^{22}\cdot405⋮405\)
a) 1.2.3.4.5.6...........10 + 324
= 6 ( a) 1.2.3.4.5.7...........10 + 54) chia hết cho 6
=> a) 1.2.3.4.5.6...........10 + 324 chia hết cho 6
b) 19.220 +76 = 19.2.110+2 . 38 = 38( 110+2) chia hết cho 38
=> ) 19.220 +76 chia hết cho 38
c) 15 . 3 . 999 + 49 = 45.999 + 45 + 4 = 45 ( 999 +1) +4 = 45 . 1000 + 4 chia 45 dư 4
=> 15 . 3 . 999 + 49 ko chia hết cho 45
Bài 1:
a. $=(-25)(-4)(-35)=100(-35)=-3500$
b. $=16-10=6$
c. $=180-(-16)-(-36)=180+16+36=232$
d. $=250-200:[1(-3)^2+(-8)]$
$=250-200:(9-8)=250-200=50$
2.
$60+2(12-x)=-48$
$2(12-x)=60-(-48)=60+48=108$
$12-x=108:2=54$
$x=12-54=-42$
Bài 1:
a. $=(-25)(-4)(-35)=100(-35)=-3500$
b. $=16-10=6$
c. $180-(-16)-(-36)=180+16+36=196+36=232$
d. $=250-200:[2000.(-3).2-6]$
$=250-200:[2000.(-6)+(-6)]$
$=250-200:[(-6)(2000+1)]=250-200[(-6).2001]$
$=250+200.6.2001=250+2401200=2401450$
Bài 2:
$60+2(12-x)=-48$
$2(12-x)=-48-60=-108$
$12-x=-108:2=-54$
$x=12-(-54)=66$
4/ Chứng minh rằng :a. 76 +75 – 74 chia hết cho 11 . bạn nào giúp mình với (giải thích cho mình hiểu luôn nha các bạ... - Hoc24
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4\cdot55⋮11\)
\(A=\left(1+7\right)+...+7^{2020}\left(1+7\right)=8\left(1+...+7^{2020}\right)⋮8\)
\(A = (1 + 7) +...+7^2\)\(^0\)\(^2\)\(^0\) \((1 + 7) = 8 (1+...+7^2\)\(^0\)\(^2\)\(^0\)\() \) ⋮\(8\)
\(A=7^3+7^4+7^5+...+7^{97}+7^{98}\)
\(=\left(7^3+7^4\right)+\left(7^5+7^6\right)+\left(7^7+7^8\right)+...+\left(7^{97}+7^{98}\right)\)
\(=7^3\left(1+7\right)+7^5\left(1+7\right)+7^7\left(1+7\right)+...+7^{97}\left(1+7\right)\)
\(=8\left(7^3+7^5+7^7+...+7^{97}\right)\) \(⋮8\) (đpcm)
Ta có :
A = 73 + 74 + 75 + 76 +........+797 + 798
=> A = 73 ( 1+ 7)+...........+797 ( 1+7)
=> A = 73 x 8 +.......+798 x 8
=> A chia hết cho 8