Tìm x: a) 2(x-2)(x+2) + 4(x-2)(x+1) + (x+2)(8-5x)+0 b) (2x+1)(5x-1) = 20x ² – 16x-1 c) 4x(2x ²-1) + 27 = (4x ²+6x+9)(2x+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}`