Question

Giải các phương trình sau:

a) \(\left( {3x + 1} \right)\left( {2 - 4x} \right) = 0;\)

b) \({x^2} - 3x = 2x - 6.\)

Answer 1

a) \(\left( {3x + 1} \right)\left( {2 - 4x} \right) = 0;\)

\(\begin{array}{l}TH1:3x + 1 = 0\\x = \frac{{ - 1}}{3}\\TH2:2 - 4x = 0\\x = \frac{1}{2}\end{array}\)

Vậy \(x \in \left\{ { - \frac{1}{3};\frac{1}{2}} \right\}\)

b) \({x^2} - 3x = 2x - 6.\)

\(\begin{array}{l}{x^2} - 3x = 2x + 6\\x\left( {x - 3} \right) = 2\left( {x + 3} \right)\\x\left( {x - 3} \right) - 2\left( {x + 3} \right) = 0\\\left( {x - 2} \right) - \left( {x - 3} \right) = 0\\TH1:x - 2 = 0\\x = 2\\TH2:x - 3 = 0\\x = 3\end{array}\)

Vậy \(x \in \left\{ {2;3} \right\}\)