15n - 3 chia het cho 15
20n - 1 chia het cho 9
18n + 3 va 21n +7
Những câu hỏi liên quan
Tìm n biết:
a) 7 chia het cho n + 1 va n > 3
b) 15 chia het cho n + 3 va 7 < n <10
c) n + 7 chia het cho n + 3
d) 2n + 6 chia het cho n + 2
ban hay giai toan nay n-6 chia het cho n+2
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a) => n+1 thuộc ước của 7
Ư(7)={-1;1;-7;7}
vì n>3 nên n=7
b) =>n+3 thuộc ước của 15
Ư(15)={-1;1;-3;3;-5;5;-15;15}
vì 7 < n < 10 nên n = 15
c) ta có : n+7 = (n+3) +4
mà n+3 chia hết cho n+3
=> 4chia hết cho n+3
=> n+3 thuôc ước của 4
Ư(4)={-1;1;-2;2;-4;4}
=> ta có bảng sau:
| n+3 | -1 | 1 | -2 | 2 | -4 | 4 |
| n | -4 | -2 | -5 | -1 | -7 | 1 |
= 2(n+2) +2d) ta có : 2n + 6 = ( 2n+4) +2
mà n+2 chia hết cho n+2 nên 2(n+2) cũng chia hết cho n+2
=> 2 phải chia hết cho n+2
=> n+2 thuộc ươc của 2
=> Ư(2)={-1;1;-2;2}
=> ta có bảng sau
| n+2 | -1 | 1 | -2 | 2 |
| n | -3 | -1 | -4 | 0 |
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a] 7 chia hết cho n+1=>n+1 thuộc Ư[7]={1;7}
=>n={0;6}
b]15 chia hết cho n+3=>n+3 thuộc Ư[15]={1;3;5;15}
=>n={0;2;4;14} mà 7<n<10=>n= rỗng
c]n+7 chia hết cho n+3
=>n+7=[n+3]+4
để [n+3]+4 chia hết cho n+3<=>4 chia hết cho n+3<=>n{1}
d]2n+6 chia hết cho n+2
có 2n+6=[n+2].2+2
mà [n+2].2 chia hết cho n+2
=>2 chia hết cho n+2<=>n={0}
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Bài 1:
a) C/m: A=2^1+2^2+2^3+2^4+....+2^2010 chia het cho 3 và 7
b) C/m: B=3^1+3^2+3^3+3^4+....+3^2010 chia het cho 4 va 13
c) C/m: C= 5^1+5^2+5^3+5^4+....+5^2010 chia het cho 6 va 31
d) C/m: D=7^1+7^2+7^3+7^4+....+7^2010 chia het cho 8 va 57
NHANH VA LUON!
Tim x
a, 35 chia het cho x + 3
b,10 chia het cho (2x + 1)
c, x + 7 chia het cho 25 va x < 100
d, x + 13 chia het cho x + 1
e, 2x + 108 chia het cho 2x + 3
a; 35 ⋮ \(x\) + 3
\(x+3\) \(\in\) Ư(35) = {-35; - 7; -5; -1; 1; 5; 7; 35}
Lập bảng ta có:
| \(x+3\) | -35 | -7 | -5 | -1 | 1 | 5 | 7 | 35 |
| \(x\) | -38 | -10 | -8 | -4 | -2 | 2 | 4 | 32 |
Theo bảng trên ta có:
\(x\in\) {-38; -10; -8; -4; -2; 2; 4; 32}
Kết luận: \(x\) {-38; -10; -8; -2; 2; 4; 32}
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b; 10 ⋮ 2\(x\) + 1
2\(x\) + 1 \(\in\) Ư(10) = {-10; -5; -2; -1; 1; 2; 5; 10}
Lập bảng ta có:
| 2\(x+1\) | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
| \(x\) | -11/2 | -3 | -3/2 | -1 | 0 | 3/2 | 2 | 11/2 |
Theo bảng trên ta có: \(x\in\) {-11/2; -3; -3/2; -1; 0; 3/2; 2; 11/2}
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c; \(x+7\) ⋮ 25
\(x+7\) \(\in\) B(25) = {0; 25; 50; 75; 100; 125;..;}
\(x\) \(\in\) {-7; 18; 43; 68; 93; 118;..;}
Vì \(x\) < 100 nên
\(x\) \(\in\) {18; 43; 68; 93}
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Xem thêm câu trả lời
chung minh rang
a tong cua 3 so chan lien tiep thi chia het cho 6
b tong cua 3 so le lien tiep khong chia het cho 6
c neu a chia het cho b va b chia het cho c thi a chia het cho c
d neu a va b chia het cho 7 co cung so du thi hieu a-b chia het cho 7
a. goi ba so tu nhien chan do la a nhan 2, a nhan 2 +2,a nhan 2 +4
theo bai ra ta co : tong ba so chan lien tiep la : a*2+a*2+2+a*2+4 = ( a*2+a*2+a*2) + (2+4)= a*6+6=6*(a+1)
vi 6 chia het cho 6 nen 6*(a+1)chia het cho 6
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nho h dung cho minh nhe cho minh nha com on
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So 35* chia het cho 2
So 1*2 chia het cho 3
So 3*5 chia het cho 9
So 1*5* chia het cho 5 va 9
So *46* chia het cho 2 ; 3; 5 va 9
Để 35* chia hết cho 2
=> * = {2 ; 4 ; 6 ; 8}
Để 1*2 chia hết cho 3
=> 1 + 2 + * chia hết cho 3
=> 3 + * chia hết cho 3
=> * = {0;3;6;9}
Để 1*5* chia hết cho 5
=> dấu * thứ 2 = {0 ; 5)
Với * thứ 2 = 0
=> 1 + * + 5 + 0 chia hết cho 9
=> 6 + * chia hết cho 9
=> * thứ 1 = 3
Với dấu sao thứ 2 = 5
=> 1 + * + 5 + 5 chia hết cho 9
=> 11 + * chia hết cho 9
=> * thứ 1 = 7
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CMR vs moi n thuoc N
a, n+2.n+7 chia het cho 2
b, 2(n+1).(n+2) chia het cho 2 va 3
c, n(n+1).(2n+1) chia het cho 2 va 3
cho abc chia het cho 3 va 7
chung ta tong ( a + 19b + 4c ) chia het cho 3 va 7
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Thay a va b bang chu so thich hop sao cho;
a) a34b chia het cho 2 , 3 va 5
b) 4a7b chia het cho 2 , 3 , 5 , 9
c) 7a8b chia het cho 5 va 9 nhung khong chia het cho 2
d) a123b chia het cho 2 , 3 va 5 nhung khong chia het cho 9
e) 47ab chia het cho 45 (2 cach )
Cho 100 so tu nhien trong do : cos 15 so khong chia het cho 2 va khong chia het cho 3 ; co 65 so chia het cho 2 va 55 so chia het cho 3 . Hoi trong 100 so do co bao nhieu so chia het cho ca 2 va 3