Question

ПОМОГИТЕ!!! СРОЧНО!!!
ПЖЖЖЖЖ!!!

Answer 1

Ответ:

1)

$1)4 {x}^{2} + 4x + 1 > 0 \\ (2x + 1) {}^{2} > 0 \\ (2x + 1) {}^{2} = 0 \\ x = - \frac{1}{2} \\ \\ 3) {x}^{2} + 49 \leqslant 14x \\ {x}^{2} + 49 - 14x \leqslant 0 \\ {x}^{2} - 14x + 49 \leqslant 0 \\ (x - 7) {}^{2} \leqslant 0 \\ (x - 7) {}^{2} = 0 \\ x = 7$