Question

Tìm số nguyên x, sao cho 7 chia hết cho x+1

Answer 1

\(7⋮\left(x+1\right)\Rightarrow\left(x+1\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ Có:x+1=-7\Rightarrow x=-8\\ x+1=-1\Rightarrow x=-2\\ x+1=1\Rightarrow x=0\\ x+1=7\Rightarrow x=6\\ Vậy:x\in\left\{-8;-2;0;6\right\}\)

Answer 2

Ta có:

\(7⋮\left(x+1\right)\Rightarrow x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có bảng sau:

\(x+1\)	\(1\)	\(-1\)	\(7\)	\(-7\)
\(x\)	\(0\)	\(-2\)	\(6\)	\(-8\)

Vậy \(x\in\left\{0;-2;6;-8\right\}\)

 

Answer 3

7⋮(x+1)⇒(x+1)∈Ư(7)={−7;−1;1;7}Coˊ:x+1=−7⇒x=−8x+1=−1⇒x=−2x+1=1⇒x=0x+1=7⇒x=6Vậy:x∈{−8;−2;0;6}