Tìm `x` biết rằng:
`a, (x^2+2x+4)(2-x)+x(x-3)(x+4)-x^2+24=0`
`=>(x^2 + 2x + 4).[-(x-2)] +(x^2 - 3x)(x+4) - x^2 + 24 = 0`
`=> - (x-2)(x^2 + 2x + 4)+(x^3 + 4x^2 - 3x^2 - 12x) - x^2 + 24 = 0`
`=> - (x^3 + 2x^2 + 4x - 2x^2 - 4x - 8)+(x^3 + 4x^2 - 3x^2 - 12x) - x^2 + 24 = 0`
`=> - (x^3 - 8) + (x^3 + 4x^2 - 3x^2 - 12x) - x^2 + 24 = 0`
`=> -x^3 + 8 + x^3 + 4x^2 - 3x^2 - 12x - x^2 + 24 = 0`
`=> -12x + 32 = 0`
`=> -12x = -32`
`=> x = (-32)/(-12)`
`=> x = 8/3`
Vậy `x = 8/3`
`b) (x/2+3)(5-6x)+(12x-2)(x/4+3)=0`
`=> (5/2 x - 3x^2 + 15 - 27x) + (3x^2 + 36x - 1/2 x - 6) = 0`
`=> 5/2 x - 3x^2 + 15 - 27x + 3x^2 + 36x - 1/2 x - 6 = 0`
`=> 20x+9 = 0`
`=> 20x = -9`
`=> x = (-9)/20`
`=> x = - 9/20`
Vậy `x = - 9/20`
