Tìm x, biết:

a) x(5 – 6x) + (2x – 1)(3x + 4) = 6;

b) x²(x – 2021) – x + 2021 = 0;

c) 2x²– 3x – 5 = 0.

Hướng dẫn giải

a) x(5 – 6x) + (2x – 1)(3x + 4) = 6

5x – 6x²+ 6x²+ 5x – 4 – 6 = 0

10x – 10 = 0

x = 1

Vậy x = 1.

b) x²(x – 2021) – x + 2021 = 0

(x – 2021)(x²– 1) = 0

(x – 2021)(x – 1)(x + 1) = 0

\( \Rightarrow \left[ \begin{array}{l}x - 2021 = 0\\x - 1 = 0\\x + 1 = 0\end{array} \right.\)

\( \Rightarrow \left[ \begin{array}{l}x = 2021\\x = 1\\x = - 1\end{array} \right.\)

Vậy x = 1, x = – 1 và x = 2021.

c) 2x²– 3x – 5 = 0

2x²+ 2x – 5x – 5 = 0

2x(x + 1) – 5(x + 1) = 0

(x + 1)(2x – 5) = 0

\(\begin{array}{l} \Rightarrow \left[ \begin{array}{l}x + 1 = 0\\2x - 5 = 0\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = - 1\\x = \frac{5}{2}\end{array} \right.\end{array}\)

Vậy x = – 1 và \(x = \frac{5}{2}\).