[tex]2x^2-3x+6 = 0 \\ \\ a)\quad x_1+x_2 = -\dfrac{b}{a} =-\dfrac{-3}{2} = \dfrac{3}{2}\\b)\quad x_1\cdot x_2 = \dfrac{c}{a} = \dfrac{6}{2} = 3 \\ c)\quad \dfrac{1}{x_1}+\dfrac{1}{x_2} = \dfrac{x_1+x_2}{x_1\cdot x_2} = \dfrac{\dfrac{3}{2}}{3} = \dfrac{1}{2}\\\\ d)\quad \dfrac{1}{x_1+3}+\dfrac{1}{x_2+3} = \dfrac{x_2+3+x_1+3}{(x_1+3)(x_2+3)} =\\ \\ = \dfrac{x_1+x_2+6}{x_1x_2+3(x_1+x_2)+9} = \dfrac{\dfrac{3}{2}+6}{3+3\cdot \dfrac{3}{2}+9} = \\ \\ = \dfrac{3+12}{24+9} = \dfrac{15}{33} = \dfrac{5}{11}[/tex]
