a) \(x\left( {x - 2} \right) = 0;\)
\(\begin{array}{l}TH1:x = 0\\TH2:x - 2 = 0\\x = 2\end{array}\)
Vậy \(x \in \left\{ {0;2} \right\}.\)
b) \(\left( {2x + 1} \right)\left( {3x - 2} \right) = 0.\)
\(\begin{array}{l}TH1:2x + 1 = 0\\x = \frac{{ - 1}}{2}\\TH2:3x - 2 = 0\\x = \frac{2}{3}\end{array}\)
Vậy \(x \in \left\{ {\frac{{ - 1}}{2};\frac{2}{3}} \right\}.\)
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