Câu 2:
Ta có: \(x^3+3x^2-4x-12=0\)
\(\Leftrightarrow x^2\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
hay \(x\in\left\{-3;2;-2\right\}\)
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