(4x + 1)(12x - 1)(3x + 2)(x + 1) = 4
[(4x + 1)(3x + 2)][(12x - 1)(x + 1)] = 4
(12x2 + 8x + 3x + 2)(12x2 + 12x - x - 1) = 4
(12x2 + 11x + 2)(12x2 + 11x - 1) = 4
Đặt 12x2 + 11x - 1 = y, ta có:
(y + 3)y = 4
\(\Rightarrow\) y2 + 3y = 4
\(\Rightarrow\) y2 + 3y - 4 = 0
\(\Rightarrow\) y2 - y + 4y - 4 = 0
\(\Rightarrow\) (y2 - y) + (4y - 4) = 0
\(\Rightarrow\) y(y - 1) + 4(y - 1) = 0
\(\Rightarrow\) (y + 4)(y - 1) = 0
\(\Rightarrow\) (12x2 + 11x - 1 + 4)(12x2 + 11x - 1 - 1) = 0
\(\Rightarrow\) (12x2 + 11x + 3)(12x2 + 11x - 2) = 0
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