`a)`\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{0;3\right\}\)
`b)`\(2x^2+2x-0,8=-\dfrac{4}{5}\)
\(\Leftrightarrow2x^2+2x=0\)
\(\Leftrightarrow2x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{-1;0\right\}\)
`c)`\(2x^2-\sqrt{4}=x-2\)
\(\Leftrightarrow2x^2-x+2-\sqrt{4}=0\)
\(\Delta=\left(-1\right)^2-4.2.\left(2-\sqrt{4}\right)\)
\(=1-16+16=1>0\)
\(\rightarrow\left\{{}\begin{matrix}x=\dfrac{1+1}{4}=\dfrac{1}{2}\\x=\dfrac{1-1}{4}=0\end{matrix}\right.\)
Vậy \(S=\left\{0;\dfrac{1}{2}\right\}\)
`d)`\(x^2+5x+6=0\)
\(\Leftrightarrow x^2+2x+3x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-2;-3\right\}\)
