Question

Решение систем неравенств. Урок 4
Срочно дам 40 баллов

Answer 1

$\displaystyle\bf\\\left \{ {{(x - 3)( x+ 5) + (x - 4) {}^{2} > 37} \atop {(x - 1)(x + 1) - (x - 2) {}^{2} \geqslant - 25 }} \right. \\ \\ 1) \: (x - 3)(x + 5) + (x - 4) {}^{2} > 37 \\ {x}^{2} + 5x - 3x - 15 + {x}^{2} - 8x + 16 - 37 > 0 \\ 2 {x}^{2} - 6x - 36 > 0 \\ {x}^{2} - 3x - 18 > 0 \\ \\ {x}^{2} - 3x - 18 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} =3 \\ x_{1} x_{2} = - 18 \\ x_{1} = - 3 \\ x_{2} = 6\\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 3x - 18 = (x + 3)(x - 6) \\ \\ (x + 3)(x - 6) > 0 \\ + + + ( - 3) - - - (6) + + + \\ x < - 3 \: \: \: and \: \: \: x > 6 \\ \\ 2) \: (x - 1)(x + 1) - (x - 2) {}^{2} \geqslant - 25 \\ {x}^{2} - 1 - {x}^{2} + 4x - 4 \geqslant - 25 \\ 4x \geqslant - 20 \\ x\geqslant - 5 \\ \displaystyle\bf\\3) \: \left \{ {{x < - 3 \: \: \: and \: \: \: x > 6} \atop {x \geqslant - 5 }} \right. \\ \\ x \: \epsilon\: [ - 5; \: - 3)U(6; \: + \propto)$