Cho A=3+3^2+3^3+3^4+...+3^100
a, tinh gia tri cua A
b, Chung to A chia het cho 10
c, tim so du cua A khi chia cho 13
d, tim x 2A-3=3^x+100
e, 2A co phai la so chinh phuong khong
dag can gap
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giup minh voi
a, 2A co phai la so chinh phuong khong
b, tim so du cua a khi chia cho 13
c, tim so du cua b khi chia cho 31
d, tim x biet ,4b + 5 = 5^x=100
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Cho B=2+2^2+.....+2^2019
a, Tim x eN sao cho A+2=2^x+100
b, A+2 co phai la mot so chinh phuong khong
c, A =3+3^2+....+3^300
chung to A chia 13
d, 2A + 3 co phai la so chinh phuong ko
a) \(A=2+2^2+....+2^{2019}\)
\(\Rightarrow2A=2^2+2^3+....+2^{2020}\)
\(\Rightarrow2A-A=2^{2020}-2\)
\(\Rightarrow A=2^{2020}-2\)
b) \(A+2=2^{2020}-2+2=2^{2020}=\left(2^{1010}\right)^2\)là SCP
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bai 1 tim cac stn x y z sao cho x+y+zxyzbai 2cmr neu m^2+mn+n^2chia het cho 9 voi m ,n la cac stn thi m,n chia het cho 3bai 3 tim stn n nho nhat de cac phan so sau toi gian7/n+9.8/n+10,...,31/n+33bai 4chung to rang phan so m^3+2m/m^4+3m^2 +1 toi gianbai 5 a,cmr khong ton tai so chinh phuong nao co dang ababb,cho 51 so nguyen duong khong vuot qua 100 cmr ton tai 2 so trong 51 so noi tren ma 1 so chia het cho so con laic,tim cac so co 3 chu so sao cho hieu cua so ay va gom 3 chu so ay viet theo th...
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bai 1 tim cac stn x y z sao cho x+y+z=xyz
bai 2cmr neu m^2+mn+n^2chia het cho 9 voi m ,n la cac stn thi m,n chia het cho 3
bai 3 tim stn n nho nhat de cac phan so sau toi gian
7/n+9.8/n+10,...,31/n+33
bai 4chung to rang phan so m^3+2m/m^4+3m^2 +1 toi gian
bai 5 a,cmr khong ton tai so chinh phuong nao co dang abab
b,cho 51 so nguyen duong khong vuot qua 100 cmr ton tai 2 so trong 51 so noi tren ma 1 so chia het cho so con lai
c,tim cac so co 3 chu so sao cho hieu cua so ay va gom 3 chu so ay viet theo thu tu nguoc lai la mot so chinh phuong
bai 6cmr A=2x(x+y-1)+y^2+1 luon nhan gia tri khong am voi moi x,y
bai 7 tim hai so nguyen duong sao cho tong ,hieu thuong (so lon chia cho so nho )cua chung cong lai duoc 38
bai 8 cho ba so a,b,c thoa man a/2009=b/2010=c/2011
tinh gia tri cua bieu thuc M=4(a-b)(b-c)-(c-a)^2
bai 1
a, chung to rang 2n+5/n+3, ( n thuoc N ) la phan so toi gian
b, tim gia tri nguyen cua n de B= 2n+5/n+3 co gia tri la so nguyen
bai 2
tim so tu nhien nho nhat sao khi chia cho 3 du 1 cho 4 du 2 cho 5 du 3 cho 6 du 4 va chia het cho 11
\(a;\frac{2n+5}{n+3}\)
Gọi \(d\inƯC\left(2n+5;n+3\right)\Rightarrow3n+5⋮d;n+3⋮d\)
\(\Rightarrow2n+5⋮d\)và \(2\left(n+3\right)⋮d\)
\(\Rightarrow\left[\left(2n+6\right)-\left(2n+5\right)\right]⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy \(\frac{2n+5}{n+3}\)là phân số tối giản
\(B=\frac{2n+5}{n+3}=\frac{2\left(n+3\right)+5-6}{n+3}=\frac{2\left(n+3\right)-1}{n+3}=2-\frac{1}{n+3}\)
Với \(B\in Z\)để n là số nguyên
\(\Rightarrow1⋮n+3\Rightarrow n+3\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow n\in\left\{-2;-4\right\}\)
Vậy.....................
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a, \(\frac{2n+5}{n+3}\)Đặt \(2n+5;n+3=d\left(d\inℕ^∗\right)\)
\(2n+5⋮d\) ; \(n+3⋮d\Rightarrow2n+6\)
Suy ra : \(2n+5-2n-6⋮d\Rightarrow-1⋮d\Rightarrow d=1\)
Vậy tta có đpcm
b, \(B=\frac{2n+5}{n+3}=\frac{2\left(n+3\right)-1}{n+3}=\frac{-1}{n+3}=\frac{1}{-n-3}\)
hay \(-n-3\inƯ\left\{1\right\}=\left\{\pm1\right\}\)
| -n - 3 | 1 | -1 |
| n | -4 | -2 |
a) tim gia tri nho nhat cua bieu thuc : A = | 1-x | + 8
b) tim cac so nguyen x biet 13 la boi cua x - 4
c) tim so nguyen x sao cho ( x + 5 ) chia het cho ( x + 3 )
giup minh voi! minh can gap gap gap.....
Bai 1: a)Tim so tu nhien a biet 1960va2002 chia cho a cung co so du la 28b)Tim 2 sop tu nhien a va b , biet :BCNN(a,b)300;UCLN(a,b)15 va a+15bBai 2:a)Tong sau la binh phuong so nao ?S1+3+5+7+...+199b) Cho so ab va so ababab 1)chung to ababab la boi cua ab2)So 3 va 10101 co phai la uoc cua ababab khong , vi sao?Bai 3a)Hay viet them dang sau so 664 ba chu so de nhan duoc sdo co 6 chu so chia het cho 5,9,11b)Tim so nguyen x thuoc Z biet rang :(x^2-1)(x^2-4)0Bai 4 :tim so nguyen x va y biet: xy-x+2...
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Bai 1: a)Tim so tu nhien a biet 1960va2002 chia cho a cung co so du la 28
b)Tim 2 sop tu nhien a va b , biet :BCNN(a,b)=300;UCLN(a,b)=15 va a+15=b
Bai 2:a)Tong sau la binh phuong so nao ?
S=1+3+5+7+...+199
b) Cho so ab va so ababab
1)chung to ababab la boi cua ab
2)So 3 va 10101 co phai la uoc cua ababab khong , vi sao?
Bai 3
a)Hay viet them dang sau so 664 ba chu so de nhan duoc sdo co 6 chu so chia het cho 5,9,11
b)Tim so nguyen x thuoc Z biet rang :
(x^2-1)(x^2-4)<0
Bai 4 :tim so nguyen x va y biet: xy-x+2y=3
Cho minh hoi bai nay. Chi can tra loi 1 cau cung duoc! Tick cho ai tra loi som nhat, day du nhat.
1. Cho so huu ti y=2a-1/-3. Voi gia tri nao cua a thi y la so duong, y la so am, khong phai so duong cung phai so am?
2. Cho so huu ti x=a-5/a (a khac 0). Voi gia tri nao cua a thi x la so nguyen?
3. Cho so huu ti x=a-3/2a (a khac 0). Voi gia tri nao cua a thi x la so nguyen?
Cho a,b thuoc Z khong la boi cua 3 nhung co cung so du khi chia cho 3. Chung to rang ab - 1 chia het cho 3.
Can gap. Giup mik nha
ta có : \(a\) có dạng \(3n+1\) hoặc \(3n+2\) và \(b\) có dạng \(3m+1\) hoặc \(3m+2\)
th1: \(a;b\) chia 3 dư \(1\) \(\Rightarrow ab-1=\left(3n+1\right)\left(3m+1\right)\)
\(=9nm+3n+3m+1-1=3\left(3nm+n+m\right)⋮3\)
th2: \(a;b\) chia 3 dư \(2\) \(\Rightarrow ab-1=\left(3n+2\right)\left(3m+2\right)\)
\(=9nm+6n+6m+4-1=3\left(3nm+2n+2m+1\right)⋮3\)
\(\Rightarrow\) đpcm
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Bai 1:tinh hop li1+2-3-4+5+6-7-8+9+10-...+2006-2007-2008+2009bai 2:cho S31+32+33+...+32014+32015a)tim so tan cung cua SB)TIM SO XBIET 3x2S+3bai3:a)cho abc chia het cho 7,chung to rang 2a+3b+c chia het cho 7b)chung to rang voi moi stn n thi 16n+3 va 12n+2 nguyen t cung nhauc) tim cac so (x;y) sao cho 34x5y chia het cho ;5;9;2d) tim stn chia cho 7 du 5;chia 13 du 4
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Bai 1:tinh hop li
1+2-3-4+5+6-7-8+9+10-...+2006-2007-2008+2009
bai 2:
cho S=31+32+33+...+32014+32015
a)tim so tan cung cua S
B)TIM SO XBIET 3x=2S+3
bai3:
a)cho abc chia het cho 7,chung to rang 2a+3b+c chia het cho 7
b)chung to rang voi moi stn n thi 16n+3 va 12n+2 nguyen t cung nhau
c) tim cac so (x;y) sao cho 34x5y chia het cho ;5;9;2
d) tim stn chia cho 7 du 5;chia 13 du 4
Bài 1
1+2-3-4+5+6-7-8+9+10-....+2006-2007-2008+2009
=1+(2-3-4+5)+(6-7-8+9)+...+(2006-2007-2008+2009)
=1+0+0+....+0
=1
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Bài 2
Ta có: S=3^1+3^2+...+3^2015
3S=3^2+3^3+...+3^2016
=> 3S-S=(3^2+3^3+...+3^2016)-(3^1+3^2+...+3^2015)
2S=3^2016-3^1
S=\(\frac{3^{2016}-3}{2}\)
Ta có \(3^{2016}=3^{4K}=\left(3^4\right)^K=\left(81\right)^K=.....1\)
=> \(S=\frac{3^{2016}-3}{2}=\frac{....1-3}{2}=\frac{....8}{2}\)
=> S có 2 tận cùng 4 hoặc 9
mà S có số hạng lẻ => S có tận cùng là 9
Ta có : 2S=3^2016-3(=)2S+3=3^2016 => X=2016
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