`19)`
`x^2 (3x+1) - (x-3)^2 = -9`
`=> (x^2 . 3x + x^2 . 1) - (x^2 - 2 . x . 3 + 3^2) = -9`
`=> (3x^3 + x^2) - (x^2 - 6x + 9) = -9`
`=> 3x^3 + x^2 - x^2 + 6x - 9 =-9`
`=> 3x^3 + 6x - 9 =-9`
`=> 3x^3 + 6x - 9 + 9 =0`
`=> 3x^3 + 6x = 0`
`=> 3x (x^2 + 2) = 0`
`=> 3x = 0 (do\ x^2 + 2 > 0 \forall x)`
`=> x= 0`
Vậy `x=0`
`20)`
`(2x+5)(4x^2 - 10x + 25) - (2x+1)^2 = 52`
`=> (2x+5) [ (2x)^2 - 2x . 5 +5^2] - [ (2x)^2 + 2 . 2x . 1 + 1^2] = 52`
`=>[ (2x)^3 + 5^3 ] - (4x^2 + 4x + 1) = 52`
`=> 8x^3 + 125 - 4x^2 - 4x -1 = 52`
`=> 8x^3 + 125 - 4x^2 - 4x -1 - 52=0`
`=> 8x^3 - 4x^2 - 4x + 72 = 0`
`=> 8x^2(x+2) - 20x(x+2) + 36(x+2) = 0`
`=> (8x^2 - 20x + 36)(x+2) =0`
`=> 4 (2x^2 - 5x + 9)(x+2) =0`
`=> (2x^2 - 5x + 9)(x+2) =0`
`=> 2x^2 - 5x + 9 =0` hoặc `x+2=0`
`+) 2x^2- 5x + 9=0`
`=> 2(x^2 - 5/2x + 25/16) + 47/8 = 0`
`=> 2 [ x^2 - 2 . x . 5/4 + (5/4)^2] + 47/8=0`
`=> 2 (x-5/4)^2 + 47/8=0`
`\forall x` ta có :
`(x-5/4)^2 \ge 0`
`=> 2(x-5/4)^2 \ge 0`
`=> 2(x-5/4)^2 + 47/8 \ge 47/8 >0`
`=> 2x^2- 5x +9 \ne 0`
`+) x + 2 = 0`
`=> x =-2`
Vậy `x=-2`
