tìm n thuoc de cc bieu thuc sau nguyen
a.A=15 CHIA HÉT CHO N-2
b.B=n-5 chia het cho n+2
c.C=2n-8 chia het cho n+2
Những câu hỏi liên quan
BAI1: CHO BIEU THUC.
A= 1494 x 1495 x 1496
KHONG THUC HIEN PHEP TINH GIAI THICH VI SAO
a; a chia het 180
b; a chia het 495
BAI2:CHUNG MINH RANG : n thuoc N thi cac so sau chia het cho 9
a; 10n-1
b; 10n+8
BAI3: TIM DIEU KIEN n thuoc N de so 10n-1 chia het cho 9 va 11
a) Vì 1494 và 1495 là số tự nhiên liên tiếp nên chia hết cho 2 , nhân với 1496 là số chẵn nên 1494 x 1495 x 1496 chia hết cho 2 => 1494 x 1495 x 1496 chia hết cho 2 x 90 => chúng chia hết cho 180.
b) Vì 1494 x 1495 x 1496 là 3 số tự nhiên liên tiếp nên chắc chắn có 1 số chia hết cho 3 => 1494 x 1495 x 1496 chia hết cho 3 => 1494 x 1495 x 1496 chia hết cho 3 x 165 => 1494 x 1495 x 1496 chia hết cho 495
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Chung minh đa thuc sau chia het cho mot so
a)n(2n-3)-2n(n+1) luon chia het cho 5 voi n thuoc Z
b)(n^2+3n-1)(n+2)-n^3+2 chia het cho 5
c)(xy-1)(x^2003+y^2003)-(xy+1)(x^2003-y^2003) chia het cho 2
a) Ta có:
\(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
\(=-5n\)
Vì \(-5n⋮5\) với n thuộc Z
\(\Rightarrow n\left(2n-3\right)-2n\left(n+1\right)⋮5\) với n thuộc Z
b) Ta có:
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n\)
\(=5\left(n^2+n\right)\)
Vì \(5\left(n^2+n\right)⋮5\)
\(\Rightarrow\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)
c) Ta có:
\(\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1-2\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1\right)\left(x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1\right)\left(x^{2003}+y^{2003}-x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)\)
\(=2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)\)
Vì \(2\left(xy+1\right)y^{2003}⋮2\)
\(2\left(x^{2003}+y^{2003}\right)⋮2\)
\(\Rightarrow2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)⋮2\)
\(\Rightarrow\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)⋮2\)
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CMR:
a,Bieu thuc : n(2n-3)-2n(n+1) luon chia het cho 5 voi moi n la so nguyen
b,Bieu thuc (n-1 )(n+4) - (n-4) (n+1)
a) \(n\left(2n-3\right)-2n\left(n+1\right)=2n^2-3n-2n^2-2n=-5n⋮5\)với \(n\inℤ\)
b) \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)=n^2+3n-4-\left(n^2-3n-4\right)=6n\)
a,CMR:Bieu thuc n(2n-3)-2n(n+1) luon chia het cho 5 voi moi n la so nguyen
b,CMR:Bieu thuc (n-1)(n+4)-(n-4)(n+10) luon chia het cho 6 voi moi so nguyen n
tim so nguyen n de
n - 2 chia het cho n + 1
2n + 7 chia het cho n + 2
+)n - 2 chia hết cho n + 1
=>n - 2 \(⋮\)n + 1
=>n + 1 - 3 \(⋮\) n + 1
Mà n + 1 \(⋮\) n + 1 nên 3 \(⋮\) n + 1
=> n + 1\(\in\)Ư(3) = {-1;1;-3;3}
=>n + 1\(\in\) {-1;1;-3;3}
=> n \(\in\){-2;0;-4;2}
Vậy n \(\in\){-2;0;-4;2}
+)2n + 7 chia hết cho n + 2
=>2n + 7 \(⋮\)n +2
=>2n + 4 +3 \(⋮\)n +2
=>2(n + 2)+ 3 \(⋮\)n + 2
Mà 2(n + 2) \(⋮\)n + 2 nên 3 \(⋮\)n + 2
=> n + 1\(\in\)Ư(3) = {-1;1;-3;3}
n + 2\(\in\) {-1;1;-3;3}
=> n \(\in\){-3;-1;-5;1}
Vậy n \(\in\){-3;-1;-5;1}
Bai 1:
a) Cho A = 963 + 351 + x voi x thuoc N . Tim dieu kien cua x de A chia het cho 9 , de A khong chia hat cho 9
b) Cho B = 10 + 25 + x + 45 voi x thuoc N . Tim dieu kien cua x De B chia het cho 5 , B khong chia het cho 5
Bai 2 : Tim x thuoc N biet :
a) 1 + 2 + 3 + ..... + n = 325
b) 1 + 3 + 5 +... + ( 2n+1) = 144
c) 2 + 4 + 6 + ... + 2n = 756
Tim n thuoc N
a) n+5 chia het cho n
b) 3n+13 chia het cho n
c) 27-5n chia het cho n
d) 2n+3 chia het cho n-2
e) 3n+1 chia het cho 11-2n
a) vi n chia het cho n nen n+5 chia het cho n khi 5 chia het cho n
do do n thuoc U(5)={1;5}
vay n=1 hoac n=5
xin loi nhe tu tu roi minh giai tiep nhe
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Cho A=n3+3n2+2n
a, CMR A chia het cho 3 voi moi so nguyen n
b, Tim gia tri nguyen duong cua n voi n<10 de A chia het cho 15
Cho A=n3+3n2+2n
a, CMR A chia het cho 3 voi moi so nguyen n
b, Tim gia tri nguyen duong cua n voi n<10 de A chia het cho 15