Решить неравенство. 7х-2≥12 2(х+1)-1<7+8х х2-5х+6>0 (х-2)(х-4)≤0

$7x - 2 \geqslant 12 \\ 7x - 14 \geqslant 0$
Найдем нули функции y = 7x - 14:
$7x = 17 \\ x = 2 \\ x \geqslant 2$

$2(x + 1) - 1 < 7 + 8x \\ 2x + 2 - 1 - 7 - 8x < 0 \\ - 6x - 6 < 0$
Найдем нули функции y = -6x - 6:
$- 6x - 6 = 0 \\ - 6x = 6 \\ x = - 1 \\ x < - 1$

0 \\ a = 1 \\ b = - 5 \\ c = 6 \\ d = {( - 5)}^{2} - 4 \times 1 \times 6 = 25 - 24 = 1 \\ \sqrt{d} = 1 \\ x1 = \frac{5 - 1}{2} = 2 \\ x2 = \frac{5 + 1}{2} = 3 " alt=" {x}^{2} - 5x + 6 > 0 \\ a = 1 \\ b = - 5 \\ c = 6 \\ d = {( - 5)}^{2} - 4 \times 1 \times 6 = 25 - 24 = 1 \\ \sqrt{d} = 1 \\ x1 = \frac{5 - 1}{2} = 2 \\ x2 = \frac{5 + 1}{2} = 3 " align="absmiddle" class="latex-formula">

$( x - 2)(x - 4) \leqslant 0 \\ {x}^{2} - 4x - 2x + 8 = 0 \\ a = 1 \\ b = - 6 \\ c = 8 \\ d = {( - 6)}^{2} - 4 \times 1 \times 8 \\ = 36 - 32 = 4 \\ \sqrt{d} = 2 \\ x1 = \frac{6 - 2}{2} = 2 \\ x2 = \frac{6 + 2}{2} = 4$