Question

tìm x biết 2x + 6 chia hết cho x + 1

Answer 1

\(2x+6⋮x+1\)

\(\Rightarrow\left\{{}\begin{matrix}2x+6⋮x+1\\x+1⋮x+1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+6⋮x+1\\2\left(x+1\right)⋮x+1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+6⋮x+1\\2x+2⋮x+1\end{matrix}\right.\)

\(\Rightarrow\left(2x+6\right)-\left(2x+2\right)⋮x+1\)

`=>2x+6 - 2x - 2` \(⋮x+1\)

\(\Rightarrow4⋮x+1\)

\(\Rightarrow x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4-4\right\}\)

Ta có từng TH sau:

`x+1=1=>x=0`

`x+1=-1=>x=-2`

`x+1=2=>x=1`

`x+1=-2=>x=-3`

`x+1=4=>x=3`

`x+1=-4=-5`

Vậy `x` $\in$ `{0;-2;1;-3;3;5}`

 

 

Answer 2

=>2x+2+4 chia hết cho x+1

=>\(x+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{0;-2;1;-3;3;-5\right\}\)