Đáp án:

Giải thích các bước giải:

`a) 6x^2 - 2x = 0`

`=> 2x(3x - 1) = 0`

`=> [(2x = 0),(3x - 1 = 0):}`

`=> [(x = 0),(3x = 1):}`

`=> [(x = 0),(x = 1/3):}`

        Vậy `x \in {0 ; 1/3}`

`b) 2x(x - 5) + 5(x - 5) = 0`

`=> (x - 5)(2x + 5) = 0`

`=> [(x - 5 = 0),(2x + 5 = 0):}`

`=> [(x = 5),(2x = -5):}`

`=> [(x = 5),(x = -5/2):}`

        Vậy `x \in {5 ; -5/2}`

`c) 2x(x - 3) + 5x - 15 = 0`

`=> 2x(x - 3) + 5(x - 3) = 0`

`=> (x - 3)(2x + 5) = 0`

`=> [(x - 3 = 0),(2x + 5 = 0):}`

`=> [(x = 3),(2x = -5):}`

`=> [(x = 3),(x = -5/2):}`

        Vậy `x \in {3 ; -5/2}`

`d) x(x - 4) - 7x + 28 = 0`

`=> x(x - 4) - 7(x - 4) = 0`

`=> (x - 4)(x - 7) = 0`

`=> [(x - 4 = 0),(x - 7 = 0):}`

`=> [(x = 4),(x = 7):}`

         Vậy `x \in {4 ; 7}`

`e) x^3 - 3x^2 - x + 3 = 0`

`=> (x^3 - 3x^2) - (x - 3) = 0`

`=> x^2(x - 3) - (x - 3) = 0`

`=> (x - 3)(x^2 - 1) = 0`

`=> [(x - 3 = 0),(x^2 - 1  = 0):}`

`=> [(x = 3),(x^2 = 1):}`

`=> [(x = 3),(x = +-1):}`

         Vậy `x \in {-1 ; 1 ; 3}`

`g) x^3 - 4x^2 + 4x = 0`

`=> x(x^2 - 4x + 4) = 0`

`=> x[(x^2 - 2x) - (2x - 4)] = 0`

`=> x[x(x - 2) - 2(x - 2)] = 0`

`=> x(x - 2)(x - 2) = 0`

`=> [(x = 0),(x - 2 = 0):}`

`=> [(x = 0),(x = 2):}`

          Vậy `x \in {0 ; 2}`