a: \(x^2\left(2x-3\right)+8x-12=0\)
=>\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(x^2+4\right)=0\)
mà \(x^2+4>=4>0\forall x\)
nên 2x-3=0
=>2x=3
=>\(x=\dfrac{3}{2}\)
b: \(2010x^2-x=2011\)
=>\(2010x^2-x-2011=0\)
=>\(2010x^2-2011x+2010x-2011=0\)
=>\(\left(2010x-2011\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}2010x-2011=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2011}{2010}\\x=-1\end{matrix}\right.\)
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