\(\text{Đặt }\left(7n+10,5n+7\right)=d\)
\(\Rightarrow\hept{\begin{cases}\left(7n+10\right)⋮d\\\left(5n+7\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left[5\left(7n+10\right)\right]d\\\left[7\left(5n+7\right)\right]⋮d\end{cases}}\)
\(\Rightarrow\left[5\left(7n+10\right)-7\left(5n+7\right)\right]⋮d\)
\(\Rightarrow1⋮d\Leftrightarrow d=1\)
\(\text{Vậy }\left(7n+10,5n+7\right)=1\)
a, 7n +10 va 5n + 7
b, 2n+3 va 4n + 8
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
a, 7n+10 va 5n+7
b, 2n+3 va 4n+8
Biet rang 3n cong 1 va 5n cong 4 (n thuoc N ) la 2 so khong nguyen to cung nhau . Tim UCLN cua 2 so tren
tim ucln cua 24 va 70
tim ucln cua 81 va 54
tim ucln cua 128 va 112
tim ucln cua 108 va 160
a) Cho a, b la 2 so nguyen to cung nhau:
a = 4n + 3; b = 5n + 1 ( n E N ). Tim UCLN( a,b).
b) Chứng minh các số sau đây nguyên tố cùng nhau: 2n + 5 va 3n + 7
1. Tim a,b thuoc so tu nhien
a. Bcnn(a,b) = 40; ucln =10
b. a - b = 48 , ucln = 12 va 10<a,b<100
c. a x b = 180, ucln = 6
d. ax b = 216, ucln = 6
e. a + b = 128, ucln = 16
2. tim ucln voi n thuoc so tu nhien (ghi cach trinh bay)
14n + 3 va 21n+ 4
tìm UCLN 7n+10 và 5n+7 với n e N
tim uc cua 24 va 70
tim uc cua 128,112 ma lon hon 10
tim uc cua 108 va 60 ma lon hon 15
tim uc cua 81 va 54
thong qua tim ucln
