Question

Giải phương trình:x^3+12=3x^2+4x

Answer 1

\(pt\Leftrightarrow x^3-3x^2-4x+12=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{-2;2;3\right\}\)

Answer 2

x³+12=3x²+4x

⇔x³-3x²-4x+12=0

⇔(x³-3x²)-(4x-12)=0

⇔x²(x-3)-4(x-3)=0

⇔(x-3)(x²-4)=0

⇔(x-3)(x-2)(x+2)=0

=>\(\left\{{}\begin{matrix}x-3=0\\x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)

vậy S{-2,2,3}