@Mốc

     2x³ + 5x² + 3x = 0

⇔ x.(2x² + 5x + 3) = 0

⇔ x.(x + 1).(2x + 3) = 0

TH1: x = 0

TH2: x + 1 = 0

 ⇔    x       = - 1

TH3: 2x + 3 = 0

⇔       x        = \(\dfrac{-3}{2}\) 

Vậy S = {0;- 1;\(\dfrac{-3}{2}\)}

      (x + 5).(x - 3) + x² - 25 = 0

⇔  (x + 5).(x - 3) + (x - 5).(x + 5) = 0

⇔  (x + 5).(x - 3 + x - 5) = 0

⇔  (x + 5).(2x - 8) = 0

TH1: x + 5 = 0

⇔     x       = - 5

TH2: 2x - 8 = 0

⇔       x      =  4

Vậy S = {- 5; 4}

     x.(x - 2) - 3x + 6 = 0

⇔ x.(x - 2) - 3.(x - 2) = 0

⇔ (x - 2).(x - 3) = 0

TH1: x - 2 = 0

⇔     x      = 2.

TH2: x - 3 = 0

⇔     x       = 3

Vậy S = {2;3}

#chucbanhoctot:)

@Mốc     

     2x + 5   = 60 - (3x - 25)

⇔ 2x + 5   = 60 - 3x + 25

⇔ 2x + 3x = - 5 + 60 + 25

⇔    5x      =        80

⇔      x      =       16

#chuchoctot:)))