Bài 1:
+ Ta có: $-3x^{2} - 5x - 2 = 0$ ⇔$\left \{ {{x_{1} = -\frac{2}{3}} \atop {x_{2} = -1}} \right.$
+ Ta có: $x_{1} + x_{2} = -\frac{2}{3} + (-1) = -\frac{5}{3}$ và $x_{1} . x_{2} = -\frac{2}{3} + (-1) = \frac{2}{3}$.
a. $M = x_{1} + \frac{1}{x_{1}} + x_{2} + \frac{1}{x_{2}}$
$M = \frac{x_{1} + x_{2}}{x_{1} . x_{2} } + (x_{1} + x_{2}) = \frac{-\frac{5}{3}}{\frac{2}{3}} - \frac{5}{3}$
$M = -\frac{5}{2} - \frac{5}{3} = \frac{-15 - 10}{6} = -\frac{25}{6}$
b. $N = \frac{1}{x_{1} + 3} + \frac{1}{x_{2} + 3}$
$N = \frac{x_{1} + x_{2} + 6}{(x_{1} + 3)(x_{2} + 3)} = \frac{(x_{1} + x_{2}) + 6}{9 + x_{1} . x_{2} + 3(x_{1} + x_{2})}$
$N = \frac{-\frac{5}{3} + 6}{9 + \frac{2}{3} + 3.(-\frac{5}{3})} = \frac{\frac{13}{3}}{\frac{14}{3}} = \frac{13}{14}$
c. $P = \frac{x_{1} - 3}{x_{1}^{2}} + \frac{x_{2} - 3}{x_{2}^{2}}$
$P = \frac{x_{2}^{2}.x_{1} - 3x_{2}^{2} + x_{1}^{2}.x_{2} - 3x_{1}^{2}}{x_{1}^{2}.x_{2}^{2}} = \frac{x_{1}x_{2}(x_{1} + x_{2}) - 3(x_{1}^{2} + x_{2}^{2})}{x_{1}^{2}.x_{2}^{2}} = \frac{x_{1} + x_{2}}{x_{1}.x_{2}} - \frac{3[(x_{1} + x_{2})^{2} - 2x_{1}.x_{2}]}{x_{1}^{2}.x_{2}^{2}}$
$P = \frac{-\frac{5}{3}}{\frac{2}{3}} - \frac{3[(-\frac{5}{3})^{2} - 2.\frac{2}{3}]}{(-\frac{5}{3}.\frac{2}{3})^{2}} = \frac{-5}{2} + \frac{\frac{25}{3} - \frac{4}{3}}{\frac{100}{81}} = \frac{713}{150}$
d. $Q = |x_{1} - x_{2}|$
$Q= \sqrt {(x_{1} - x_{2})^{2}} = \sqrt {(x_{1} + x_{2})^{2} - 4x_{1}.x_{2}}$
$Q = \sqrt {(-\frac{5}{3})^{2} - 4.\frac{2}{3}} = \frac{1}{3}$
