\(\Leftrightarrow3x-6+13⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;13;-13\right\}\)
hay \(x\in\left\{3;1;15;-11\right\}\)
3x+7⋮ x+2
3(x+2)+1⋮x+2
1⋮x+2
x+2∈Ư(1)={±1}
| x+2 | 1 | -1 |
| x | -1 | -3 |
⇔3x−6+13⋮x−2⇔3x−6+13⋮x−2
⇔x−2∈{1;−1;13;−13}⇔x−2∈{1;−1;13;−13}
hay x∈{3;1;15;−11}
\(\left(3x+7\right)⋮\left(x-2\right)\\ \Rightarrow\left[\left(3x-6\right)+13\right]⋮\left(x-2\right)\\ \Rightarrow\left[3\left(x-2\right)+13\right]⋮\left(x-2\right)\)
Vì \(3\left(x-2\right)⋮\left(x-2\right)\Rightarrow13⋮\left(x-2\right)\Rightarrow x-2\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\)
Ta có bảng:
| x-2 | -13 | -1 | 1 | 13 |
| x | -11 | 1 | 3 | 15 |
Vậy \(x\in\left\{-11;1;3;15\right\}\)
\(\left(3x+7\right)⋮\left(x-2\right)\\ 3x-2+9⋮x-2\\ 3x-2⋮x-2\\ \Rightarrow9⋮x-2\\ \Rightarrow x-2\in\text{Ư}\left(9\right)=\left\{-1;-3;-9;1;3;9\right\}\\ x-2=-1\\ x=\left(-1\right)+2\\ x=1\\ x-2=-3\\ x=\left(-3\right)+2\\ x=-1\\ x-2=-9\\ x=\left(-9\right)+2\\ x=-7\\ x-2=1\\ x=1+2\\ x=3\\ x-2=9\\ x=9+2\\ x=11\\ x-2=3\\ x=3+2\\ x=5\)
vậy......
