Giải thích các bước giải:
Ta có : `2x + 3 = 2x + 8 - 8 + 3`
`=> 2x + 8 - 8 + 3 \vdots x+4`
`=> 2(x+4) -5 \vdots x+4`
Mà `2(x+4) \vdots x+4 => -5 \vdots x+4`
`=> x+4 in Ư{-5}`
`=> x+4 in {+- 5 ; +-1 }`
`=> x in {1;-3;-5;-9}`
Vậy `x in {1;-3;-5;-9}`
`b)`
Ta có : `3x - 5 = 3x + 6 - 6 - 5`
`=> 3x + 6 - 6 - 5 \vdots x+2`
`=> 3(x+2) - 11 \vdots x+2`
Mà `3(x+2) \vdots x+2 => 11 \vdots x+2`
`=> x+2 in Ư{11}`
`=> x+2 in {+- 1 ; +- 11}`
`=> x in {-1;-3;-13;9}`
Vậy `x in {-1;-3;-13;9}`
`c)` Ta có : `7 - 4x = 7 - 4x - 6 +6`
`=> 7 - 4x - 6 + 6 \vdots 2x+3`
`=> (7+6) - 4x - 6 \vdots 2x+3`
`=> 13 - 4x - 6 \vdots 2x+3`
`=> 13 - 2.(2x+3) \vdots 2x+3`
Mà `2(2x+3) \vdots 2x+3 => 13 \vdots 2x+3`
`=> 13 \vdots 2x+3`
`=> 2x + 3 in Ư(13)`
`=> 2x +3 in {+- 1 ; +-13}`
`=> 2x in {-2;-4;10;-16}`
`=> x in {-1;-2;5;-8}`
Vậy `x in {-1;-2;5;-8}`
