\(1,\)
\(\left\{{}\begin{matrix}2x+y=10\\5x-3y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+5y=50\\10x-6y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+5y-10x+6y=50-6\\2x+y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11y=44\\2x+y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\2x+4=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
Vậy hệ pt có tập nghiệm là \(\left(x;y\right)=\left(3;4\right)\)
\(2,3x^2-2x-1=0\)
\(\Delta=b^2-4ac=\left(-2\right)^2-4.3.\left(-1\right)=16>0\)
\(\Rightarrow\)Pt có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{2+4}{6}=1\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{2-4}{6}=-\dfrac{1}{3}\end{matrix}\right.\)