Question

Câu 1 :

a) ( x + 12 ) \(⋮\) ( 10 - x )

b) 2^{x + 1} - 12 = 3² + 2017⁰

c) x - 3 \(⋮\) ( x + 1 )

d) ( 2x + 1 ) \(⋮\) x - 2

Answer 1

***********************************************************

a) Ta có: \(\dfrac{x+12}{10-x}=-\dfrac{x-10+22}{x-10}=-1+\dfrac{22}{x-10}\)

Vì \(\left(x+12\right)⋮\left(10-x\right)\) nên \(22⋮\left(x-10\right)\)

Do đó ta có bảng:

x-10	-22	-11	-2	-1	1	2	22
x	-12	-1	8	9	11	12	32

Vậy \(x\in\left\{-12;-1;8;9;11;12;32\right\}\)

Answer 2

c) \(\left(x-3\right)⋮\left(x+1\right)\)

=> \(\left(x-3\right)-\left(x+1\right)⋮\left(x+1\right)\)

=> \(\left(x-3-x-1\right)⋮\left(x+1\right)\)

=>\(-4⋮\left(x+1\right)\)

=> x+1\(\in\) ư(-4)= \(\left\{\pm1,\pm2,\pm4\right\}\)

ta có bảng sau

x+1	-4	-2	-1	1	2	4
x	-5	-3	-2	0	1	3

vậy x\(\in\left\{-5,-3;-2;0;1;3\right\}\)