Tìm x biết:
a) \({x^2} - 4x = 0\)
b) \(2{x^3} - 2x = 0\)
Phân tích đa thức thành nhân tử.
\(A.B = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{A = 0}\\{B = 0}\end{array}} \right.\)
a)
\(\begin{array}{l}{x^2} - 4x = 0\\ \Leftrightarrow x\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x - 4 = 0}\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x = 4}\end{array}} \right.\end{array}\)
Vậy \(x \in \left\{ {0;4} \right\}\)
b)
\(\begin{array}{l}2{x^3} - 2x = 0\\ \Leftrightarrow 2x\left( {{x^2} - 1} \right) = 0\\ \Leftrightarrow 2x\left( {x - 1} \right)\left( {x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x - 1 = 0\\x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 1\\x = - 1\end{array} \right.\end{array}\)
Vậy \(x \in \left\{ {0;1; - 1} \right\}\)