\[\boxed{\text{217\ (217).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ 2x^{2} - 10x + 3 = 0\]
\[D = 5^{2} - 2 \cdot 3 = 19 > 0 \Longrightarrow\]
\[\Longrightarrow существует\ 2\ корня,\ \]
\[по\ теореме\ Виета:\]
\[2x^{2} - 10x + 3 = 0\ \ \ \ |\ :2\]
\[x^{2} - 5x + 1,5 = 0\]
\[x_{1} \cdot x_{2} = 1,5;\ \ \ \ \ \ \ \ x_{1} + x_{2} = 5.\]
\[\textbf{б)}\frac{1}{3}x^{2} + 7x - 2 = 0\]
\[x^{2} + 21x - 6 = 0\]
\[D = 21^{2} + 4 \cdot 6 > 0\]
\[\Longrightarrow существует\ 2\ корня,\ \]
\[по\ теореме\ Виета:\]
\[\frac{1}{3}x^{2} + 7x - 2 = 0\ \ \ \ | \cdot 3\]
\[x^{2} + 21x - 6 = 0\]
\[x_{1} \cdot x_{2} = - 6;\ \ \ \ \ \ \ \ x_{1} + x_{2} = - 21.\]
\[\textbf{в)}\ 0,5x^{2} + 6x + 1 = 0\ \ \ \ \ \ \ \ | \cdot 2\]
\[x^{2} + 12x + 2 = 0\]
\[D = 6^{2} - 2 = 36 - 2 = 34 > 0\]
\[\Longrightarrow существует\ 2\ корня,\ \]
\[по\ теореме\ Виета:\]
\[x_{1} \cdot x_{2} = 2;\ \ \ \ \ \ \ x_{1} + x_{2} = - 12;\]
\[\textbf{г)} - \frac{1}{2}x^{2} + \frac{1}{3}x + \frac{1}{2} = 0\ | \cdot ( - 2)\]
\[x^{2} - \frac{2}{3}x - 1 = 0\]
\[D = \frac{4}{9} + 4 \cdot 1 > 0\]
\[\Longrightarrow существует\ 2\ корня,\ \]
\[по\ теореме\ Виета:\]
\[x_{1} \cdot x_{2} = \frac{2}{3};\ \ \ \ \ \ x_{1} + x_{2} = - 1.\ \]