Bài 4 b) (x+2)(x-3)+ x - 1 = 0 c) (3x + 1).(2x - 3) - 6x. (x+2) = 0 d) (2x-1) (3x + 1) +(4 - 2x)3x = 7

`# Ly`

`b)`

       `(x+2)(x-3)+x-1=0`

`⇔ x^2 -x -6 +x-1=0`

`⇔ x^2 - 7=0`

`⇔ x^2 = 7`

`⇔ x^2 = (sqrt7)^2`

`⇔ x= ± sqrt7`

Vậy `S={sqrt7 ; - sqrt7}`

`c)`

         `(3x+1)(2x-3) - 6x(x+2)=16`

`⇔ 6x^2 - 9x + 2x - 3  - (6x^2 + 12x)=16`

`⇔ 6x^2 - 7x -3 - 6x^2 - 12x - 16=0`

`⇔ - 19x -19=0`

`⇔ -19x= 19`

`⇔ x= -1`

Vậy `S={-1}`

`d)`

              `(2x-1)(3x+1)+ (4-2x). 3x= 7`

`⇔ 6x^2 + 2x -3x -1 + 12x - 6x^2 -7=0`

`⇔ 11x -8=0`

`⇔ 11x= 8`

`⇔ x= 8/11`

Vậy `S={8/11}`