Question

Знайти область визначення

Answer 1

$\displaystyle\bf y = \frac{x + 3}{ \sqrt{14 - 3x - 2 {x}^{2} } } + \frac{x - 1}{2 {x}^{2} - 3x + 1} \\ \left \{ {{14 - 3x - 2 {x}^{2} > 0 } \atop {2 {x}^{2} - 3x + 1 \neq0}} \right. \\ \\ - 2 {x}^{2} - 3x + 14 > 0 \\ 2 {x}^{2} + 3x - 14 < 0 \\ 2 {x}^{2} + 3x - 14 = 0 \\ a =2 \\ b = 3 \\ c = - 14 \\ D = {b}^{2} - 4ac = {3}^{2} - 4 \times 2 \times ( - 14) = \\ = 9 + 112 = 121 \\ x_{1} = \frac{ - 3 - 11}{2 \times 2} = - \frac{14}{4} = - 3.5 \\ x_{2} = \frac{ - 3 + 11}{2 \times 2} = \frac{8}{4} = 2 \\ 2 {x}^{2} + 3x - 14 = 2(x + 3.5)(x - 2) \\ (x + 3.5)(x - 2) < 0 \\ + + + ( - 3.5) - - - (2) + + + \\ x \: \epsilon \: ( - 3.5; \: 2) \\ \\ {2x}^{2} - 3x + 1 = 0 \\ a = 2 \\ b = - 3 \\ c =1 \\ D = {b}^{2} - 4ac = ( - 3) {}^{2} - 4 \times 2 \times 1 = \\ = 9 - 8 = 1 \\ x_{1} = \frac{3 - 1}{2 \times 2} = \frac{2}{4} = 0.5\\ x_{2} = \frac{3 + 1}{2 \times 2} = \frac{4}{4} = 1 \\ x\neq0.5 \: \: \: \: \: and \: \: \: \: \: x\neq1 \\ \\ \displaystyle\bf\\\left \{ {{x \: \epsilon \: ( - 3.5; \: 2)} \atop {x \neq0.5 \: \: \: \: \: and \: \: \: \: \: x\neq1}} \right. \\ \\ otvet \: \: \: x \: \epsilon\: ( - 3.5; \: 0.5)U(0.5; \: 1)U(1; \: 2)$