Question

8-2x=корень x+1 прошу помогите
И
log2(1-2x)<0

Answer 1

$(8 - 2x) {}^{2} = ( \sqrt{x + 1} ) {}^{2}$
$64 - 36x + 4{x}^{2} - x - 1 = 0$
$4 {x}^{2} - 37x + 63 = 0$
$d = 1369 - 4 \times 4 \times 63 = \\ = 1369 - 1008 = 361$
$d = \sqrt{361} = 19$
$x = \frac{ 37 + 19}{8} = 7$
$x = \frac{37 - 19}{8} = 2 \frac{2}{8}$

Ответ: (2 2/8; 7)

$log_{2}(1 - 2x) < 0$
$log_{2}(1 - 2x) < log_{2}(2) \\ 1 - 2x < 2 \\ - 2x < 2 - 1 \\ x > - \frac{1}{2}$