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MV
26 tháng 6 2017
\(A=17^{18}-17^{16}\\ =17^{16}\cdot\left(17^2-1\right)\\ =17^{16}\cdot\left(289-1\right)\\ =17^{16}\cdot288\\ =17^{16}\cdot18\cdot16⋮18\)
Vậy \(A⋮18\)
\(B=1+3+3^2+...+3^{11}\)
Ta có: \(52=4\cdot13\)
\(B=1+3+3^2+...+3^{11}\\ =\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\\ =1\cdot\left(1+3\right)+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\\ =\left(1+3\right)\cdot\left(1+3^2+...+3^{10}\right)\\ =4\cdot\left(1+3^2+...+3^{10}\right)⋮4\)
Vậy \(B⋮4\)
\(B=1+3+3^2+...+3^{11}\\ =\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\\ =1\cdot\left(1+3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^9\cdot\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\cdot\left(1+3^3+...+3^9\right)\\ =13\cdot\left(1+3^3+...+3^9\right)⋮13\)
Vậy \(B⋮13\)
Vì \(4\) và \(13\) là hai số nguyên tố cùng nhau nên tao có \(B⋮4\cdot13\Leftrightarrow B⋮52\)
Vậy \(B⋮52\)
\(C=3+3^3+3^5+...3^{31}\)
\(C=3+3^3+3^5+...+3^{31}\\ =\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\\ =1\cdot\left(3+3^3\right)+3^4\cdot\left(3+3^3\right)+...+3^{28}\cdot\left(3+3^3\right)\\ =\left(3+3^3\right)\cdot\left(1+3^4+...+3^{28}\right)\\ =30\cdot\left(1+3^4+...+3^{28}\right)⋮15\left(\text{vì }30⋮15\right)\)
Vậy \(C⋮15\)
\(D=2+2^2+2^3+...+2^{60}\)
Tao có: \(21=3\cdot7;15=3\cdot5\)
\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)
Vậy \(D⋮3\)
\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\\ =2\cdot\left(1+2^2\right)+2^5\cdot\left(1+2^2\right)+...+2^{57}\cdot\left(1+2^2\right)+2^2\cdot\left(1+2^2\right)+...+2^{58}\cdot\left(1+2^2\right)\\ =\left(1+2^2\right)\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)\\ =5\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)⋮5\)
Vậy \(D⋮5\)
\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)
Ta có:
\(D⋮3;D⋮5\Rightarrow D⋮3\cdot5\Leftrightarrow D⋮15\)
\(D⋮3;D⋮7\Rightarrow D⋮3\cdot7\Leftrightarrow D⋮21\)
Vậy \(D⋮15;D⋮21\)
26 tháng 6 2017
Mình chỉ làm mẫu 1 câu thui nha:
\(A=17^{18}-17^{16}\)
\(A=17^{16}.17^2-17^{16}.1\)
\(A=17^{16}\left(17^2-1\right)\)
\(A=17^{16}.288\)
\(A=17^{16}.16.18\)
\(A⋮18\left(đpcm\right)\)
MD
2
11 tháng 12 2017
a) |-18| . 941 + 59 . 18
= 18 . 941 + 59 . 18
= ( 941 + 59 ) . 18
= 1000 . 18
= 18000
b) 81 : 33 + 16 : 23
= 81 : 27 + 16 : 8
= 3 + 2
= 5
c) 30 - [ 40 - ( 6 - 1)2]
= 30 - [ 40 - 52 ]
= 30 - [ 40 - 25 ]
= 30 - 15
= 15
d) 17 . 85 + 15 . 17 - 150
= ( 85 + 15 ) . 17 - 150
= 100 . 17 - 150
= 1700 - 150
= 1550
QD
11 tháng 12 2017
a, \(\left|-18\right|\cdot941+59\cdot18=18\cdot941+59\cdot18=18\cdot\left(941+59\right)=18\cdot1000=18000\)
b, \(81:3^3+16:2^3=81\cdot27+16\cdot8=2187+128=2315\)
c, \(30-\left[40-\left(6-1\right)^2\right]=30-\left[40-5^2\right]=30-\left[40-25\right]=30-15=15\)
d, \(17\cdot85+15\cdot17-150=17\cdot\left(85+15\right)-150=17\cdot100-150=1700-150=1550\)
HN
1
CC
5 tháng 12 2019
\(A=3+3^2+3^3+.......+3^{17}\)
\(\Rightarrow3A=3^2+3^3+3^4+......+3^{18}\)
\(\Rightarrow3A-A=2A=3^{18}-3\)\(\Rightarrow A=\frac{3^{18}-3}{2}\)
Vì \(-3>-4\)\(\Rightarrow3^{18}-3>3^{18}-4\)\(\Rightarrow A>B\)
MV
23 tháng 4 2017
a) Không thể vì: \(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}=1+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>1\)
b) Ta có: \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
CM: \(\dfrac{a}{b}=\dfrac{a\cdot\left(b-m\right)}{b\cdot\left(b-m\right)}=\dfrac{ab-am}{b^2-bm}\left(1\right)\\ \dfrac{a-m}{b-m}=\dfrac{\left(a-m\right)\cdot b}{\left(b-m\right)\cdot b}=\dfrac{ab-am}{b^2-bm}\left(2\right)\)
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow am< bm\Rightarrow ab-am>ab-bm\left(3\right)\)
Từ (1), (2), (3) ta có \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
Vậy
\(B=\dfrac{17^{19}-1}{17^{20}-1}>\dfrac{17^{19}-1-16}{17^{20}-1-16}=\dfrac{17^{19}-17}{17^{20}-17}=\dfrac{17\cdot\left(17^{18}-1\right)}{17\cdot\left(17^{19}-1\right)}=\dfrac{17^{18}-1}{17^{19}-1}=A\)
Vậy B > A
10 tháng 6 2015
a.17 . 85 + 25 . 17 + 1200
=17.(85+25)+1200
=17.110+1200
=1870+1200
=3070
b.62 : 4 . 3 + 2 . 52
=36:4.3+2.25
=27+50
=77
c.5 . 42 - 18 : 32
=5.16-18:9
=80-2
=78
d.3 . 52 - 16 : 23 + 34 : 33
=3.25-16:8+3
=75-2+3
=76
e.220 - [ 32 . 33 - ( 12 - 70 ) 2]
=220-35+112
=220-243+121
=98
j. 49 . 73 + 49 . 51 - 49 . 24
=49.(73+51-24)
=49.100
=4900
Con bấm mỏi suốt đấy mấy má
MT
10 tháng 6 2015
nhầm
a) 17 . 85 + 25 . 17 + 1200
=(17.85+25.17)+1200
=17(85+25)+1200
=17.110+1200
=1870+1200
=3070
b) 62 : 4 . 3 + 2 . 52
=36:4.3+2.25
=9.3+50
=27+50
=77
c) 5 . 42 - 18 : 32
=5.16-18:9
=80-2
=78
d) 3 . 52 - 16 : 23 + 34 : 33
=3.25-24:23+3(4-3)
=75-2(4-3)+31
=75-21+3
=75-2+3
=76
e) 220 - [ 32 . 33 - ( 12 - 70 ) 2 ]
=220-[3(2+3)-(12-1)2]
=220-(35-112)
=220-(243-121)
=220-122
=98
j) 49 . 73 + 49 . 51 - 49 . 24
=49(73+51-24)
=49.100
=4900
232 nha bn
@Mina
232 nha bạn. Ko cần k.