a)
\(\left(x-15\right)⋮\left(x+2\right)\)
\(\Rightarrow x+2-17⋮\left(x+2\right)\)
\(\Rightarrow-17⋮\left(x+2\right)\Rightarrow\left(x+2\right)\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3;15;-19\right\}\)
b)
\(\left(3x+16\right)⋮\left(x+1\right)\)
\(\Rightarrow3.\left(x+1\right)+13⋮\left(x+1\right)\)
\(\Rightarrow13⋮\left(x+1\right)\Rightarrow\left(x+1\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;-14;12\right\}\)
a) x - 15 là Bội của x+ 2 => x - 15 chia hết cho x+ 2
Ta có x-15=(x+2)-17=> (x+2)-17 chia hết cho x+ 2
mà (x+2) chia hết cho x+ 2 =>17 chia hết cho x+ 2
=>x+2 ∈ Ư(17)
x+2 ∈ {-17:-1;1;17}
| x+2 | x |
| -17 | -19 |
| -1 | -3 |
| 1 | -1 |
| 17 | 15 |
