Giải bất phương trình: \(\frac{{{{\left( {x - 1} \right)}^3}.{{\left( {x - 2} \right)}^2}}}{{\left| {3x + 4} \right|}} \ge 0\) ta được:
A.\(\left[ \begin{array}{l}x \ge 1\\x = 2\end{array} \right.\)
B.\(\left[ \begin{array}{l}x \ge 2\\x = - \frac{4}{3}\end{array} \right.\)
C.\(x \ge 1\)
D.\(\left[ \begin{array}{l}x = - \frac{4}{3}\\x \ge 1\end{array} \right.\)
A.\(\left[ \begin{array}{l}x \ge 1\\x = 2\end{array} \right.\)
B.\(\left[ \begin{array}{l}x \ge 2\\x = - \frac{4}{3}\end{array} \right.\)
C.\(x \ge 1\)
D.\(\left[ \begin{array}{l}x = - \frac{4}{3}\\x \ge 1\end{array} \right.\)
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Hóa Học lớp 12
