(1,5 điểm):

Tìm x biết:

a) 15x²– 3x = 0;

b) (3x – 2)(x + 3) + (x²– 9) = 0;

c) (x – 1)³– (x + 1)(2 – 3x) = – 3.

Hướng dẫn giải

a) 15x²– 3x = 0

⇔ 3x(5x – 1) = 0

\( \Leftrightarrow \left[ \begin{array}{l}3x = 0\\5x - 1 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = \frac{1}{5}\end{array} \right.\)

Vậy x = 0, \(x = \frac{1}{5}\).

b) (3x – 2)(x + 3) + (x²– 9) = 0

⇔ (3x – 2)(x + 3) + (x + 3)(x – 3) = 0

⇔ (x + 3)(3x – 2 + x – 3) = 0

⇔ (x + 3)(4x – 5) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x + 3 = 0\\4x - 5 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = - 3\\x = \frac{5}{4}\end{array} \right.\)

Vậy x = – 3; \(x = \frac{5}{4}\).

c) (x – 1)³– (x + 1)(2 – 3x) = – 3

⇔ x³– 3x²+ 3x – 1 + 3x²+ x – 2 + 3 = 0

⇔ x³+ 4x = 0

⇔ x(x²+ 4) = 0

⇔ x = 0 (vì x²+ 4 >0 với mọi x)

Vậy x = 0.