`2) (2x + 1)^2 - (x-1)^2 = 0`

`[(2x + 1) - (x-1)][(2x + 1) + (x-1)] = 0`

`[2x + 1 - x + 1][2x + 1 + x - 1] = 0`

 `[x + 2]3x = 0`

`x (x + 2) = 0`

`x = 0` hoặc `x + 2 =0`

                    `x = -2`

Vậy `x in {-2;0}`

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`3) (3x - 5)^2 - (x+1)^2 = 0`

`[(3x - 5) - (x+1)][(3x - 5) + (x+1)] = 0`

`[3x - 5 - x - 1][3x - 5 + x + 1] = 0`

`[2x  - 6][4x - 4] = 0`

`2x - 6= 0` hoặc `4x - 4 = 0`

`2x = 6`                 `4x = 4`

`x = 3`                    `x  =1`

Vậy `x in {1;3}`