Question

УМАЛЯЮ СРОЧНО ДАМ 100 БАЛОВ ВСЕ ЧТО УГОДНО ​

Answer 1

4.

$\displaystyle\bf\\\left \{ {{35 - 2x - { {x}^{2} > 0}^{} } \atop {5x + 1 \leqslant - 1 - 5x }} \right. \\ \\ 1) \: 35 - 2x - {x}^{2} > 0 \\ {x}^{2} + 2x - 35 < 0 \\ \\ {x}^{2} + 2x - 35 = 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} = - 2 \\ x_{1} x_{2} = - 35 \\ x_{1} = - 7\\ x_{2} =5 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} + 2x - 35 = (x + 7)(x - 5) \\ \\ (x + 7)(x - 5) < 0 \\ + + + ( - 7) - - - (5) + + + \\ - 7 < x < 5 \\ \\ 2) \: 5x + 1 \leqslant - 1 - 5x \\ 5x + 5x \leqslant - 1 - 1 \\ 10x \leqslant - 2 \: \: | \div 10 \\ x \leqslant - 0.2 \\ \displaystyle\bf\\3) \: \left \{ {{ - 7 < x < 5} \atop {x \leqslant - 0.2 }} \right. \\ \\ otvet \: \: \: x \: \epsilon \: ( - 7; \: - 0.2]$

5.

${x}^{2} - 14x + 48= 0 \\ po \: \: \: teoreme \: \: \: vieta \\ x_{1} + x_{2} = 14 \\ x_{1} x_{2} = 48\\ x_{1} = 6 \\ x_{2} = 8 \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2} \\ {x}^{2} - 14x + 48 = (x - 6)(x - 8) \\ \\ \frac{ {x}^{2}(x - 1) }{ {x}^{2} - 14x + 48} \leqslant 0 \\ \frac{ {x}^{2}(x - 1) }{(x - 6)(x - 8)} \leqslant 0 \\ \\ \left \{ {{ {x}^{2} (x - 1)(x - 6)(x - 8) \leqslant 0} \atop {x \neq6 \: \: \: \: \: \: \: \: x\neq8}} \right. \\ \\ - - - [0] - - - [1] + + + (6) - - - (8) + + + \\ x \: \epsilon\: ( - \propto; \: 1]U(6; \: 8)$

6.

$\displaystyle\bf\\\left \{ {{ {x}^{2} + 4x + 3 \leqslant 0 } \atop {2 {x}^{2} + 5x < 0}} \right. \\ \\ 1) \: {x}^{2} + 4x + 3 \leqslant 0 \\ {x}^{2} + 4x + 3 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ x_{1} + x_{2} = - 4 \\ x_{1} x_{2} = 3 \\ x_{1} = - 3 \\ x_{2} = - 1 \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ x {}^{2} + 4x + 3 = (x + 3)(x + 1) \\ (x + 3)(x + 1) \leqslant 0 \\ + + + [ - 3] - - - [ - 1] + + + \\ - 3 \leqslant x \leqslant - 1 \\ \\ 2) \: 2 {x}^{2} + 5x < 0 \\ {x}^{2} + 2.5x < 0 \\ x(x + 2.5) < 0 \\ + + + ( - 2.5) - - - (0) + + + \\ - 2.5 < x < 0 \\ \displaystyle\bf\\3) \: \left \{ {{ - 3 \leqslant x \leqslant - 1} \atop { - 2.5 < x < 0 }} \right. \\ \\ x \:\epsilon \: ( - 2.5; \: - 1]$