`a)2(x-2)(x+2)+4(x-2)(x+1)+(x+2)(8-5x)=0`

`⇔2(x²-4)+(4x-8)(x+1)+8x-5x²+16-10x=0`

`⇔2x²-8+4x²+4x-8x-8+8x-5x²+16-10x=0`

`⇔(2x²+4x²-5x²)+(4x-8x+8x-10x)+(-8-8+16)=0`

`⇔x²-6x=0`

`⇔x(x-6)=0`

`(1)x=0`

`(2)x-6=0⇔x=6`

Vậy `x=0` hoặc `x=6`

`b)(2x+1)(5x-1)=20x²-16x-1`

`⇔10x²-2x+5x-1=20x²-16x-1`

`⇔10x²-2x+5x-20x²+16x=-1+1`

`⇔(10x²-20x²)+(-2x+5x+16x)=0`

`⇔-10x²+19x=0`

`⇔x(-10x+19)=0`

`(1)x=0`

`(2)-10x+19=0⇔x=19/10`

Vậy `x=0` hoặc `x=19/10`

`c)4x(2x²-1)+27=(4x²+6x+9)(2x+3)`

`⇔8x³-4x+27=8x³+12x²+18x+12x²+18x+27`

`⇔8x³-4x-8x³-12x²-18x-12x²-18x=27-27`

`⇔(8x³-8x³)+(-12x²-12x²)+(-4x-18x-18x)=0`

`⇔-24x²-40x=0`

`⇔-8x(3x+5)=0`

`(1)-8x=0⇔x=0`

`(2)3x+5=0⇔x=(-5)/3`

Vậy `x=0` hoặc `x=(-5)/3`

`a) 2(x - 2)(x + 2) + 4(x - 2)(x + 1) + (x + 2)(8 - 5x) = 0`

`=> 2(x^2 - 2^2) + 4(x^2 + x - 2x - 2) + 8x -  5x^2 + 16 - 10x = 0`

`=> 2x^2 - 8 + 4x^2 + 4x - 8x - 8 + 8x - 5x^2 + 16 - 10x = 0`

`=> (2x^2 + 4x^2 - 5x^2) + (4x - 8x + 8x - 10x) + (-8 - 8 + 16) = 0`

`=> x^2 - 6x = 0`

`=> x(x - 6) = 0`

`=>` \(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\) 

Vậy `x in {0; 6}`

`b) (2x + 1)(5x - 1) = 20x^2 - 16x - 1`

`=> 10x^2 - 2x + 5x - 1 = 20x^2 - 16x - 1`

`=> 10x^2 + 3x - 1 = 20x^2 - 16x - 1`

`=> 20x^2 - 16x - 1 - 10x^2 - 3x + 1 = 0`

`=> (20x^2 - 10x^2) - (16x + 3x) + (-1 + 1) = 0`

`=> 10x^2 - 19x = 0`

`=> x(10x - 19) = 0`

`=>` \(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{19}{10}\end{array} \right.\) 

Vậy `x in {0; 19/10}`

`c) 4x(2x^2 - 1) + 27 = (4x^2 + 6x + 9)(2x + 3)`

`=> 8x^3 - 4x + 27 = 8x^3 + 12x^2 + 12x^2 + 18x + 18x + 27`

`=> 8x^3 + 12x^2 + 12x^2 + 18x + 18x + 27 - 8x^3 + 4x - 27 = 0`

`=> (8x^3 - 8x^3) + (12x^2 + 12x^2) + (18x + 18x + 4x) + (27 - 27) = 0`

`=> 24x^2 + 40x = 0`

`=> x(24x + 40) = 0`

`=>` \(\left[ \begin{array}{l}x=0\\24x+40=0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{-5}{3}\end{array} \right.\) 

Vậy `x in {0; (-5)/3}`