Question

с сором по алгебре, от 1. Решите неравенства:
+3x+6<0
b) -r² +5x-16>0
2. Решите систему неравенств:
[18-3x <0;
x²-x-12 20.
[5]
[5]
[6]

Answer 1

1.

а)

$\frac{1}{3} {x}^{2} + 3x + 6 < 0 \: \: | \times 3 \\ {x}^{2} + 9x + 18 < 0 \\ {x}^{2} + 9x + 18 = 0$

По теореме Виета :

${x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c$

$x_{1} + x_{2} = -9 \\ x_{1} x_{2} = 18\\ x_{1} = - 6 \\ x_{2} = - 3 \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} + 9x + 18 = (x + 3)(x + 6) \\ \\ (x + 3)(x + 6) < 0 \\ + + + ( - 6) - - - ( - 3) + + + \\ - 6 < x < - 3 \\ x \: \epsilon \: ( - 6; \: - 3)$

б)

$- {x}^{2} + 5x - 16 0 \\ {x}^{2} - 5x + 16 < 0 \\ {x}^{2} - 5x + 16 = 0 \\ a =1 \\ b = - 5\\ c = 16\\ D = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 1 \times 16 = 25 - 64 = - 39 \\ D < 0 \\ x \: \epsilon \: \varnothing$

2.

$\left \{ {{18 - 3x < 0} \atop { {x}^{2} - x - 12 \geqslant 0 }} \right. \\ \\ 1) \: 18 - 3x < 0 \\ - 3x < - 18 \: \: | \div ( - 3) \\ x 6 \\ \\ 2) \: {x}^{2} - x - 12 \geqslant 0 \\ {x}^{2} - x - 12 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ x_{1} + x_{2} = 1\\ x_{1} x_{2} = - 12 \\ x_{1} = - 3 \\ x_{2} = 4 \\ ( x + 3)(x - 4) \geqslant 0 \\ + + + [ - 3] - - - [4] + + + \\ x \leqslant - 3 \: \: \: and \: \: \: x \geqslant 4 \\ \\ 3) \: \left \{ {{x 6} \atop {x \leqslant - 3 \: \: \: and \: \: \: x \geqslant 4}} \right. \\ \\ x \: \epsilon \: (6; \: + \infty)$