Tìm x: a) 8x^3 – 12x^2 + 6x – 1 = 0; b) (4x – 3)^2 – 3x(3 – 4x) = 0.

Trả lời

a) 8x³ – 12x² + 6x – 1 = 0

$\iff$(2x)³ – 3.(2x)².1 + 3. 2x. 1² - 1³ = 0

$\iff$(2x – 1)³ = 0

$\iff$2x – 1 = 0

$\iff x=\frac{1}{2}$.

Vậy $x=\frac{1}{2}$.

b)

(4x – 3)² – 3x(3 – 4x) = 0

$\iff$(4x – 3)² + 3x(4x – 3) = 0

$\iff$(4x – 3)(4x – 3 + 3x) = 0

$\iff$(4x – 3)(7x – 3) = 0

$\iff \left[\begin{array}{l}4x-3=0\\ 7x-3=0\end{array}\right.$

$\iff \left[\begin{array}{l}x=\frac{3}{4}\\ x=\frac{3}{7}\end{array}\right.$

Vậy $x=\frac{3}{4}$ hoặc $x=\frac{3}{7}$.