Question

tìm x \(\in\)N

x+10\(⋮\)x+1

Answer 1

Ta có:

\(x+10⋮x+1\)

\(\Rightarrow\left(x+1\right)+9⋮x+1\)

\(\Rightarrow9⋮x+1\)

\(\Rightarrow x+1\in U\left(9\right)=\left\{1;3;6;9\right\}\) (vì x là số tự nhiên )

+) \(x+1=1\Rightarrow x=0\)

+) \(x+1=3\Rightarrow x=2\)

+) \(x+1=6\Rightarrow x=5\)

+) \(x+1=9\Rightarrow x=8\)

Vậy x=0 ; x=2 ; x=5 ; x=8

Answer 2

Ta có: x + 10 = (x + 1) + 9 \(\Rightarrow\) (x + 1) + 9 \(⋮\) (x + 1) khi 9 \(⋮\) (x + 1)

\(\Rightarrow\) x + 1 \(\in\) Ư(9) = {1; 3; 9}

\(\Rightarrow\) x \(\in\) {0; 2; 8}

Vậy x \(\in\) {0; 2; 8}.