Question

решите систему неравенств 8x-2<x-1
2x^2-x-1<=0
​

Answer 1

1.

$8x - 2 < x - 1 \\ 8x - x < - 1 + 2 \\ 7x <1 \\ x < \frac{1}{7} \\ (- \infty \: \: \frac{1}{7} )$

2.

${2x}^{2} - x - 1 \leqslant 0 \\ d = ( { - 1}^{2} ) - 4 \times 2 \times ( - 1) = 1 + 8 = 9 \\ x 1= \frac{1 + \sqrt{9} }{2 \times 2} = \frac{4}{4} = 1 \\ x2 = \frac{1 - \sqrt{9} }{2 \times 2} = \frac{ - 2}{4} = - 0.5 \\ | - 0.5 \: .\: 1 |$