a) \(2x\left(x-1\right)-\left(x-2\right)\left(x+2\right)-\left(x-3\right)^2=3\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x^2-4\right)-\left(x^2-6x+9\right)-3=0\)
\(\Leftrightarrow2x^2-2x-x^2+4-x^2+6x-9-3=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=8:4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(x^3+x^2-9x-9=0\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(9x+9\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x=-1\) hoặc \(x=3\) hoặc \(x=-3\)
