\(x^2+x-7⋮x+1\)
\(\Rightarrow x^2-1+x-6=\left(x+1\right)\left(x-1\right)+x-6⋮x+1\)
Vì \(\left(x+1\right)\left(x-1\right)⋮\left(x+1\right)\Rightarrow\left(x-6\right)⋮x+1\)
\(x-6=\left(x+1\right)-7⋮x+1\)
Do đó \(x+1|7\)
Xét các TH
x+1 = ( 1,7,-1,-7)
=> x=(0;6;-2;-8)
làm linh tinh thôi
Ta có \(x^2+x-7⋮x+1\)
\(\Leftrightarrow x\left(x+1\right)-7⋮x+1\)
\(\Leftrightarrow7⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Leftrightarrow x\in\left\{-8;-2;0;6\right\}\) ( thỏa mãn x nguyên )
Vậy \(x\in\left\{-8;-2;0;6\right\}\)
@@ K chắc lắm nha
Học tốt
## Takigawa Miu_
\(x^2+x-7=x^2+2x+1-x-1-7=\left(x+1\right)^2-\left(x+1\right)-7\)
Do \(\left(x+1\right)^2-\left(x+1\right)⋮\left(x+1\right)\)
\(\Rightarrow x^2+x-7⋮x+1\Leftrightarrow-7⋮\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\inƯ\left(-7\right)=\left(\pm1;\pm2;\pm7\right)\)
\(\Rightarrow x\in\left\{0;-2;1;-3;6;-8\right\}\)
Vậy ...