Question

помогите пожалуйста!!

Answer 1

$\displaystyle\bf\\y = \sqrt{3x + 9} - \frac{ {x}^{2} - 1}{ {x}^{2} + 2x - 15} \\ \left \{ {{3x + 9 \geq 0 \: \: | \div 3} \atop { {x}^{2} + 2x - 15\neq0}} \right. \\ \\ {x}^{2} + 2x - 15 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ x_{1} + x_{2} = - 2 \\ x_{1} x_{2} = - 15\\ x_{1} = - 5 \\ x_{2} = 3 \\ \\ \left \{ {{x \geq - 3} \atop {x\neq - 5 \: \: \: and \: \: \: x\neq3 }} \right. \\ \\ otvet \: \: \: x \:\epsilon \: [ - 3; \: 3)U(3; \: + \propto)$