a ) Ta có : x + 5 \(⋮\)x + 2
\(\Leftrightarrow\)( x + 2 ) + 3 \(⋮\)x + 2
\(\Leftrightarrow\)x + 2 \(\in\)Ư( 3 ) = { \(\pm\)1 ; \(\pm\)3 }
Ta lập bảng :
| x + 2 | 1 | - 1 | 3 | - 3 |
| x | - 1 | - 3 | 1 | - 5 |
Vậy : ...............
b ) Tương tự nhé .
a. x+5 chia hết cho x+2
<=> x+2+3 chia hết cho x+2
<=> 3 chia hết cho x+2
=> x+2 \(\in\)Ư(3)={-1,-3,1,3}
| x+2 | -1 | -3 | 1 | 3 |
| x | -3 | -5 | -1 | 1 |
Vậy.....
b. x+4 chia hết cho x-2
<=> x-2+6 chia hết cho x-2
<=> 6 chia hết cho x-2
=> x-2 \(\in\)Ư(6)={-1,-2,-3,-6,1,2,3,6}
| x-2 | -1 | -2 | -3 | -6 | 1 | 2 | 3 | 6 | |
| x | 1 | 0 | -1 | -4 | 3 | 4 | 5 | 8 |
Vậy.....
a, \(\frac{x+5}{x+2}\)=\(\frac{\left(x+2\right)+3}{x+2}\)= \(1+\frac{3}{x+2}\)
Để \(\frac{x+5}{x+2}\)\(\in\) \(z\)thì \(\frac{3}{x+2}\)\(\in\)\(z\)\(\Rightarrow\)\(x+2\in\)Ư(3)
Ta có bảng sau:
| x+2 | 1 | 3 | -1 | -3 |
| x | -1 | 1 | -3 | -5 |
Vậy \(x\in\left\{-1;1;-3;-5\right\}\)
\(\in\)\(\left\{-1;1;-3;-5\right\}\)}
b, \(\frac{x+4}{x-2}\)=\(\frac{\left(x-2\right)+6}{x-2}\)=\(1+\frac{6}{x-2}\)
Để \(\frac{x+4}{x-2}\)\(\in\) \(z\)thì \(\frac{6}{x-2}\)\(\in\) \(z\)\(\Rightarrow\)\(x-2\in\)Ư(6)
Ta có bảng sau:
| x-2 | 1 | 2 | 3 | 6 | -1 | -2 | -3 | -6 |
| x | 3 | 4 | 5 | 8 | 1 | 0 | -1 | -4 |
Vậy \(x\in\)\(\left\{3;4;5;8;1;0;-1;-4\right\}\)
chúc bn hok tốt !
