Ta có \(x+5⋮x+2\)
=> x + 2 + 3 \(⋮\)x + 2
Vì x + 2 \(⋮\)x + 2
=> 3 \(⋮\)x + 2
=> x + 2 \(\in\)Ư(3)
=> x + 2 \(\in\) {1 ; 3 - 1 ; - 3}
=> x \(\in\){-1 ; 1 ; - 3 ; - 5}
b) (x - 2)(x + 3) = 0
\(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
=> x \(\in\){2 ; - 3}
c) (2x + 60)(9 - x2) = 0
=> \(\orbr{\begin{cases}2x+60=0\\9-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-30\\x=\pm3\end{cases}}\)
=> x \(\in\){- 30 ; 3 ; - 3}
d) Vì \(x;y\inℤ\Rightarrow\hept{\begin{cases}x-2\inℤ\\y+1\inℤ\end{cases}}\)
Ta có 3 = 1.3 = (-1).(-3)
Lập bảng xét các trường hợp
| x - 2 | 1 | - 1 | 3 | - 3 |
| y + 1 | 3 | - 3 | 1 | - 1 |
| x | 3 | 1 | 5 | -1 |
| y | 2 | -4 | 0 | -2 |
Vậy các cặp số (x ; y) thỏa mãn là (3 ; 2) ; (1 ; - 4) ; (5 ; 0) ; (- 1; - 2)
a, \(x+5⋮x+2\)
\(\Leftrightarrow x+2+3⋮x +2\)
\(\Leftrightarrow3⋮x+2\) => \(x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
b, \(\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
c, \(\left(2x+60\right)\left(9-x^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+60=0\\9-x^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-60\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-30\\x=\pm3\end{cases}}}\)
d, \(\left(x-2\right)\left(y+1\right)=3\Rightarrow x-2;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| y + 1 | 3 | -3 | 1 | -1 |
| x | 3 | 1 | 5 | -1 |
| y | 2 | -4 | 0 | -2 |
