a) \(x^2 + x = 6 \)
\(\Rightarrow x^2+x-6=0\)
\(\Rightarrow x^2-2x+3x-6=0\)
\(\Rightarrow\left(x^2-2x\right)+\left(3x-6\right)\)\(=0\)
\(\Rightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x= 2\) ; \(x = -3\)
b) \(6x^3 + x^2 = 2x\)
\(\Rightarrow6x^3+x^2-2x=0\)
\(\Rightarrow6x^3-3x^2+4x^2-2x=0\)
\(\Rightarrow\left(6x^3-3x^2\right)+\left(4x^2-2x\right)=0\)
\(\Rightarrow3x^2\left(2x-1\right)+2x\left(2x-1\right)=0\)
\(\Rightarrow x\left(2x-1\right)\left(3x+2\right)=0\)
\(\Leftrightarrow x=0\)
Hoặc \(2x - 1 = 0 \) \(\Rightarrow2x=1\) \(\Rightarrow x=\dfrac{1}{2}\)
Hoặc \(3x + 2 = 0 \) \(\Rightarrow3x=-2\) \(\Rightarrow x=\dfrac{-2}{3}\)
Vậy \(x = 0\) ; \(x=\dfrac{1}{2}\) ; \(x=-\dfrac{2}{3}\)
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