Câu ` 1 : `

Ta có ` : `

` ( 2x+ 6 )/( 2x + 3 ) `

` = ( 2x + 3 + 3 )/( 2x + 3 ) `

` = 3/( 2x + 3 ) `

Để ` ( 2x + 6 ) \vdots ( 2x + 3 ) ` thì ` : `

` ( 2x + 6 )/( 2x + 3 ) ` là số nguyên 

` => ( 2x + 3 ) in Ư(3) `

Mà ` : Ư(3) = { -3 ; -1 ; 1 ; 3 } `

TH1 ` : `

` 2x + 3 = -3 `

` 2x = -6 `

` x = -3 `

` `

TH2 ` : `

` 2x + 3 = -1 `

` 2x = -4 `

` x = -2 `

` `

TH3 ` : `

` 2x + 3 = 1 `

` 2x = -2 `

` x = -1 ` 

` `

TH4 ` : `

` 2x + 3 = 3 `

` 2x = 0 `

` x = 0 `

` => x in { -3 ; -2 ; -1 ; 0 } ` để ` ( 2x + 6 ) \vdots ( 2x + 3 ) `

Vậy ` : ... `

` text{#Antanezha} `