Question

4. Решите систему неравенств: Поште неравенство - x² + 2x + 8 ≤ 0 - ( 6 - 2 (x + 1) > 2x​

Answer 1

Ответ:

Объяснение:

{ -x² + 2x + 8 ≤ 0    | *(-1)

{ (6 - 2 (x + 1) > 2x​

{ x² - 2x - 8 ≥ 0

{ (6 - 2x - 2 > 2x​

{(х - 4) * (х + 2) ≥ 0

{4х < 4

х ≥ 4  х ≤ -2                              х < 1

х ∈ ( -∞; - 2] ∪ [ 4; + ∞)           х ∈ ( -∞; 1)

Ответ:  х ∈ ( -∞; - 2]

Answer 2

$\displaystyle\bf\\\left \{ {{ - {x}^{2} + 2x + 8 \leqslant 0 } \atop {6 - 2(x + 1) > 2x }} \right. \\ \\ 1) \: - {x}^{2} + 2x + 8 \leqslant 0 \\ {x}^{2} - 2x - 8 \geqslant 0 \\ {x }^{2} - 2x - 8 = 0 \\po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} =2 \\ x_{1} x_{2} = - 8 \\ x_{1} =4 \\ x_{2} = - 2 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 2x - 8 = (x - 4)(x + 2) \\ \\ (x - 4)(x + 2) \geqslant 0 \\ + + + [ - 2] - - - [4] + + + \\ x \leqslant - 2 \: \: \: and \: \: \: x \geqslant 4 \\ \\ 2) \: 6 - 2(x + 1) > 2x \\ 6 - 2x - 2 - 2x> 0 \\ - 4x > - 4 \: \: | \div ( - 4) \\ x < 1 \\ \displaystyle\bf\\3) \: \left \{ {{x \leqslant - 2 \: \: \: and \: \: \: x \geqslant 4} \atop {x < 1 }} \right. \\ \\ otvet \: \: \: x \:\epsilon \: ( - \propto; \: - 2]$