cho a, b la 2 so nguyen khong la boi cua 3 nhung co cung so du khi chia cho 3.Chung to rang (ab-1) chia het cho 3
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
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 a,b la hai so nguyen ko chia het cho 3 nhung khi chia cho 3 thi co cung so du.CMR ab-1 la bởi cua 3
Cho so a= 36.q + 15 voi q la so tu nhien
A. Chung to rang a khong chia het cho 2
B. Chung to rang a chia het cho 3 va a khong phai la so nguyen to
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.....................
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 |
KHI CHIA MOT SO A CHO 12 DUOC SO DU LA 9.Chung tỏ rang a khong chia het cho 3 nhung a khong chia het cho 4
bai 1
cho 2 so nguyen a,b ko chia het cho 3 nhung khi chia cho 3 thi co cung so du chung minh ab-1 la boi cua 3 ( goi y ap dung cong thuc (a+b).(c+d)=a.(c+d)+b.(c+d) )
bai 2 thuc hien phep tinh M=1+2+22+23+.....+22012/22014-2
b, cho S=5+52+53+54+55+56+.........+52012 chung to S chia het cho 65
giup minh voi can ban oi!!!!!!!!!!!!!!!
1, a,b ko chia hết cho 3 nhưng có cùng số dư khi chia cho 3
=> a,b cùng chia 3 dư 1 hoặc 2
sau đó xét 2 TH;
=> ab chia 3 dư 1 => ab-1 là bội của 3 (ĐPCM)
Ta có:
S=1+2+2^2+.......+2^2012
2S=(2+2^2+2^3+........+2^2013)
S=2^2013-1=(2^2014-2)/2
=> S=1/2
Câu b tra con nhà bà mạng :D
b1 tim so tu nhien lon nhat co 3 chu so biet rang chia cho 5; 7 ; 9 thi co so du lan luot la 2 ; 4 ;6
b2 tim x biet
a ) 2x +3 chia het cho x -1
b)3x+5 chia het cho x+1
b3 chung to rang 3n+4 va 4n +5 la 2 so nguyen to cung nhau vs moi so TN
Goi y
B1 X+3 chia het cho 5 7 9
B2 a ; Nhan x-1 vs 2 Roi tru cho nhau
b ; nhan x+1 vs 3
B3 nhan 3n +4 vs 4 ; 4n +5 vs3 roi tru
hai so khong chia het cho 3 khi cho thi duoc nhung so du khac nhau chung to rang to cua hai so do chia het cho 3