Question

tìm số tự nhiên n biết:

â, n+9 chia hết cho n+2

b, 2n+9 chia hết cho n-1

c, 3n+5 chia hết cho 2n+1

Answer 1

\(a,n+9⋮n+2.\)

\(\Rightarrow\left(n+2\right)+7⋮n+2.\)

mà \(n+2⋮n+2\Rightarrow7⋮n+2\Rightarrow n+2\in U_{\left(7\right)}=\left\{1;7\right\}.\)

\(\Rightarrow n\in\left\{-1;-9\right\}.\)

Vậy..........

\(b,2n+9⋮n-1.\)

\(\Rightarrow\left(2n-2\right)+11⋮n-1.\)

\(\Rightarrow2\left(n-1\right)+11⋮n-1.\)

mà \(2\left(n-1\right)⋮n-1\Rightarrow11⋮n-1\Rightarrow n-1\in U_{\left(11\right)}=\left\{1;11\right\}.\)

\(\Rightarrow n\in\left\{2;12\right\}.\)

Vậy..........

\(c,3n+5⋮2n+1.\)

\(\Rightarrow2\left(3n+5\right)⋮2n+1.\)

\(\Rightarrow6n+10⋮2n+1.\)

\(\Rightarrow\left(6n+3\right)+7⋮2n+1.\)

\(\Rightarrow3\left(2n+1\right)+7⋮2n+1.\)

mà \(3\left(2n+1\right)⋮2n+1\Rightarrow7⋮2n+1\Rightarrow2n+1\in U_{\left(7\right)}=\left\{1;7\right\}.\)

\(\Rightarrow n\in\left\{0;3\right\}.\)

Vậy..........

Answer 2

a) n+9⋮n+2= (n+2)+7⋮n+2

=> n+2 ∈ Ư(7)={1;7}

ĐK: n ∈ N

Nếu n+2=1⇒n=1-2=-1 (vì -1∉ N⇒loại)

n+2=7⇒n=7-2=5 (vì 5 ∈ N⇒chọn)

Vậy n=5

b)2n+9⋮n-1=(n-1)+(n-1)+11⋮n-1

⇒ n-1 ∈ Ư(11)={1;11}

ĐK: n∈ N

Nếu n-1=1⇒n=1+1=2 (vì 2∈N⇒chọn)

n-1=11⇒n=11+1=12 ( vì 12 ∈ N⇒chọn)

Vậy n={2;12}

c)3n+5⋮2n+1=(n+1)+(n+1)+(n+1)+2⋮(n+1)+n

ĐK: n∈ N

⇒ n ∈ Ư(2)={1;2}

⇒ n=1(thỏa mãn)

n=2(thỏa mãn)

Vậy n={1;2}