Tim 2 so a va b biet UCLN(a,b)=12;BCNN(a,b)=72
1,tim 2 so tu nhien biet tong chung = 66 UCLN=6 co 1 so chia het cho 5
2,tim 2 so tu nhien biet hieu chung = 84 UCLN = 12
3,tim 2 so tu nhien biet tich = 864 UCLN=6
4,cho 3n+1 va 5n+4 la 2 so nguyen to cung nhau tim UCLN cua 3n+1 va 5n+4
5,cho a=123456789;b=987654321 . Tim UCLN cua a va b
Tim 2 so tu nhien a,b biet tong cua chung bang 60 va UCLN (a,b) =12.
Biet rang : BCNN(a,b).UCLN(a,b)
a, BCNN cua hai so la 600, UCLN cua chung nho hon 10 lan BCNN. So thu 1 la 120, tim so thu 2.
b, UCLN cua hai so la 12, BCNN cua chung lon gap 6 lan UCLN . So t 1 la 24, tim so t 2
c, Tong cua hai so bang 60, tong giua UCLN va BCNN cua chung la 84. Tim hai so do.
Tim 2 so a va b, biet a>b;a+b=16 va UCLN(a,b)=4
tim hai so a va b biet a.b=2250va UCLN(a,b)=15
tin 2 so a va b biet rang a.b=1176;BCNN(a,b)=84 va a>b
giai giup minh nhe
bai 1 a )tim ucln(6n+3,6n+9)
b)tim ucln (2n-1;9n+4)
bai2 tim a;b thuoc n biet
bcnn(a;b)=240
ucln(a;b)=16
bai 3 tim so chia va thuong cua 1 phep chia biet so bi chia la 145 du 12 thuong # 1
Tim hai so tu nhien a va b biet
a>b;a+b=108 va UCLN (a,b)=12
tim so tu nhiên a va b biet ucln (a va ) = 4 va a + b = 48
Lời giải:
Vì $ƯCLN(a,b)=4$ nên đặt $a=4x, b=4y$ với $x,y$ là số tự nhiên, $x,y$ nguyên tố cùng nhau.
Ta có:
$a+b=4x+4y=48$
$x+y=48:4=12$
Vì $x,y$ nguyên tố cùng nhau nên $(x,y)=(1,11), (5,7), (7,5), (11,1)$
$\Rightarrow (a,b)=(4,44), (20, 28), (28,20), (44,4)$
tim 2 so a;b biet UCLN(a;b)=6 va a.b=864