A) 4X(X + 1) + (3 – 2X)(3 + 2X) = 15⇔4X2 + 4X + (9 – 4X2) = 15⇔ 4...

Bài 2a) 4x(x + 1) + (3 – 2x)(3 + 2x) = 15⇔4x

²

+ 4x + (9 – 4x

²

) = 15⇔ 4x

²

+ 4x + 9 – 4x

²

= 15⇔4x = 15 – 9⇔4x = 6⇔x = 3/2b)3x(x – 20012) – x + 20012 = 0⇔3x(x – 20012) – (x – 20012) = 0⇔(x – 20012)(3x – 1) = 0⇔x – 20012 = 0 hay 3x – 1 = 0⇔x = 20012 hoặc x = 1/2