a)(2x-3)(4x+1) - 8x(x-2) = 12

<=> 8x² - 10x - 3 - (8x² - 16x) = 12

<=> 6x - 3 = 12

<=> 6x = 15 <=> x = 5/2

b)(2x+3)² - (2x-3)(2x+3) = 8

<=> 4x² + 12x + 9 - (4x² - 9) = 8

<=> 12x + 18 = 8

<=> 12x = -10 <=> x = -5/6

(12x-5)(4x-1) + (3x-7)(1-16x) = 81

<=>(48x² - 32x + 5) + (-48x² + 115x - 7) = 81

<=>83x - 2 = 81

<=>83x = 83 <=> x=1

Vậy x=1

Ta có (x-y)³-(y-z)³-(z-x)³

= x³-3x²y+3xy²-y³+y³-3y²z+3yz²-z³+z³-3z²x+3zx²-x³

= -3x²y+3xy²-3y²z+3yz²-3z²x+3zx²

= 3(xy²+yz²+zx²-x²y-y²z-z²x)

= 3[(y-x)xy+(z-y)yz+(x-z)zx]

Ta có SABCD = (10,6+5,5).6:2 = 48 cm²

SBCE = 5,4.6:2 = 16,2 cm²

=> SABCD-SBCE = 48-16,2 = 31,8 cm²