Question

Знайдіть область визначения функції (На фото)

Answer 1

$\displaystyle\bf y = \sqrt{20 + x - {x}^{2} } + \frac{4}{x - 2} \\ \left \{ {{ - {x}^{2} + x + 20 \geqslant 0} \atop {x - 2\neq0 }} \right. \\ \\ 1) \: - {x}^{2} + x + 20 \geqslant 0 \\ {x {}^{2} - x - 20} \leqslant 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ {x}^{2} - x - 20 = 0 \\ x_{1} + x_{2} = 1 \\ x_{1} x_{2} = - 20 \\ x_{1} = 5\\ x_{2} = - 4 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - x - 20 = (x - 5)(x + 4) \\ \\ (x - 5)(x + 4) \leqslant 0 \\ + + + [ - 4] - - - [5] + + + \\ - 4 \leqslant x \leqslant 5 \\ \\ 2) \: x - 2\neq0 \\ x\neq2 \\ \\ \displaystyle\bf\\\left \{ {{ - 4 \leqslant x \leqslant 5} \atop {x \neq2}} \right. \\ \\ otvet \: \: \: x \:\epsilon \: [ - 4; \: 2)U(2; \: 5]$