tim n thuoc N sao cho
n.(n+6) la so nguyen to
n^2+10.n la so nguyen to
cho n thuoc N* biet n-10,n+10,n+60 la so nguyen to . CMn+90 la so nguyen to
Bai1 a,cho n thuoc N. Chung minh rang 6n+5 va 4n+3 la 2 so nguyen to cung nhau
b, tim so nguyen x sao cho x+2016 la so nguyen duong nho nhat
Bài 1
a,
Gọi d là ƯCLN(6n+5;4n+3)
\(\Rightarrow\hept{\begin{cases}6n+5⋮d\\4n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(6n+5\right)⋮d\\3\left(4n+3\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}12n+10⋮d\\12n+9⋮d\end{cases}}}\)
\(\Rightarrow12n+10-\left(12n+9\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow\) d=1 hay ƯCLN (6n+5;4n+3) =1
Vậy 6n+5 và 4n+3 là 2 số nguyên tố cùng nhau
b, Vì số nguyên dương nhỏ nhất là số 1
=> x+ 2016 = 1
=> x= 1-2016
x= - 2015
Đặt \(6n+5;4n+3=d\left(d\inℕ^∗\right)\)
\(6n+5⋮d\Rightarrow12n+10⋮d\)
\(4n+3⋮d\Rightarrow12n+9⋮d\)
Suy ra : \(12n+10-12n-9⋮d\)hay \(1⋮d\)
Vậy ta có đpcm
tim m;n thuoc N sao cho m^2+2 la so nguyen to va 2m^2=n^2_2
tim n thuoc Z de so p=10/n^2+4 la so nguyen to chan
3n + 1.
tim n thuoc N sao cho 3n + 1 la so nguyen to
a, tim so nguyen to P+10; p+20 cung la so nguyen to
b, tim UCLN(3n+2;4-1);(a thuoc N)
tim n thuoc N de 2^n-1 va 2^n+1 la so nguyen to
tim cac so nguyen duong n sao cho n^2/60-n la 1 so nguyen to
tim so tu nhien n nho nhat sao cho n+2,n+6 la so nguyen to