`a)`
`(5x-1)(x+1) - 2 (x+3)^2 = 3x (x+7)`
`=> (5x . x + 5x . 1 - 1 . x -1 . 1) - 2 (x^2 + 2 . x . 3 + 3^2) = 3x . x + 3x . 7`
`=> (5x^2 + 5x - x - 1) - 2 (x^2 + 6x + 9) = 3x^2 + 21x`
`=> 5x^2 + 5x - x - 1 - 2x^2 - 12x - 18 = 3x^2 + 21x`
`=> 5x^2 + 5x - x - 1 - 2x^2 - 12x - 18 - 3x^2 - 21x =0`
`=> (5x^2 - 2x^2 - 3x^2) + (5x - x - 12x - 21x) - (1+18) = 0`
`=> -29x - 19 = 0`
`=> -29x = 19`
`=> x = -19/29`
Vậy `x=-19/29`
`b)`
`2 (x-3)^2 - (x-3)(x+3) = (x-4)^2`
`=> 2 (x^2 - 2 . x . 3 + 3^2) - (x^2 - 3^2) = x^2 - 2 . x . 4 + 4^2`
`=> 2 (x^2 - 6x + 9) - (x^2 - 9) = x^2 - 8x + 16`
`=> 2x^2 - 12x + 18 - x^2 + 9 = x^2 - 8x + 16`
`=> 2x^2 - 12x + 18 - x^2 + 9 - x^2 + 8x - 16 =0`
`=> (2x^2 - x^2 - x^2) + (8x - 12x) + (18+ 9-16) = 0`
`=> - 4x + 11 =0`
`=> -4x = -11`
`=> x = 11/4`
Vậy `x=11/4`
`c)`
`-5 (x+3)^2 + (x-1)(x+1) = (3-2x)^2`
`=> -5 (x^2 + 2 . x . 3 + 3^2) + (x^2 - 1) = 3^2 - 2 . 3 . 2x + (2x)^2`
`=> -5 (x^2 + 6x + 9) + (x^2 - 1) = 9 - 12x + 4x^2`
`=> -5x^2 - 30x - 45 + x^2 - 1 = 9 - 12x + 4x^2`
`=> -5x^2 - 30x - 45 + x^2 - 1 - 9 + 12x - 4x^2 = 0`
`=> (-5x^2 + x^2 - 4x^2) + (12x - 30x) - (1+9+45) = 0`
`=> -8x^2 -18x - 55=0`
`=> 8x^2 + 18x + 55 = 0`
`=> 8 (x^2 + 9/4x + 81/64) + 359/8 =0`
`=> 8 [ x^2 + 2 . x . 9/8 + (9/8)^2] + 359/8 =0`
`=> 8 (x+9/8)^2 + 359/8 = 0` (không xảy ra)
Vậy không tìm được `x` thỏa mãn yêu cầu đề bài.
