\(\left\{{}\begin{matrix}5x+2y=12\\3x-2y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x+2y+3x-2y=12+4\\3x-2y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}8x=16\\3x-2y=4x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\3\cdot2-2y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
1)\(A=4\sqrt{12}+5\sqrt{48}-3\sqrt{108}=4\sqrt{2^2.3}+5\sqrt{4^2.3}-3\sqrt{6^2.3}\)
\(A=8\sqrt{3}+20\sqrt{3}-18\sqrt{3}=10\sqrt{3}\)
3)`x^2+x-6=0`
`<=>(x-2)(x+3)=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 `S={2;-3}`
