Ta có:
\(2x+9⋮x-2\)
\(\Rightarrow\left(2x-4\right)+13⋮x-2\)
\(\Rightarrow2\left(x-2\right)+13⋮x-2\)
\(\Rightarrow13⋮x-2\)
\(\Rightarrow x-2\in\left\{-1;1;-13;13\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=-1\Rightarrow x=1\\x-2=1\Rightarrow x=3\\x-2=-13\Rightarrow x=-11\\x-2=13\Rightarrow x=15\end{matrix}\right.\)
Vậy x=1 hoặc x=3 hoặc x=-11 hoặc x=15
Ta có: 2x + 9 = 2(x - 2) + 13
Vì 2(x- 2) \(⋮\) (x-2) => 13 \(⋮x-2\Rightarrow x-2\inƯ\left(13\right)\)
\(\Rightarrow x-2\in\left\{\pm1;\pm13\right\}\)
\(\Rightarrow x\in\left\{3;1;15;-11\right\}\).
(2x+9) ⋮ (x-2)
vì (x-2)⋮(x-2)
=> 2(x-2)⋮(x-2)
=> (2x-4)⋮(x-2)
=> (2x+9)-(2x-4)⋮(x-2)
=> (2x+9-2x+4)⋮(x-2)
=> 13⋮(x-2)
=> x-2 ∈ Ư(13)=\(\left\{\pm1;\pm13\right\}\)
ta có bảng sau
| x-2 | -13 | -1 | 1 | 13 |
| x | -11 | -1 | 3 | 11 |
vậy x∈{-11;-1;3;11}