Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!!

РЕШИТЕ ПОКАЗАТЕЛЬНЫЕ НЕРАВЕНСТВА

ДАЮ 60 баллов!!!!

Answer 1

Ответ:

$1) {2}^{4(7- 5x)} = {2}^{ - 3(2x + 1)} \\ 28 - 20x = - 6x - 3 \\ - 14x = - 31 \\ x = \frac{31}{14}$

$2) {3}^{5} \times {3}^{ - 3(3x - 2)} = {3}^{3(x + 3)} \\ 5 - 3(3x - 2) = 3(x + 3) \\ 5 - 9x + 6 = 3x + 9 \\ - 12x = - 2 \\ x = \frac{1}{6}$

$3)2 \times {5}^{x} \times {5}^{2} - 10 \times {5}^{x} = 8 \\ {5}^{x} (2 \times {5}^{2} - 10) = 8 \\ {5}^{x} \times 40 = 8 \\ {5}^{x} = \frac{1}{5} \\ x = - 1$

$4)4 \times {3}^{x + 2} + 5 \times {3}^{x + 1} - 6 \times {3}^{x} = 5 \\ {3}^{x} (4 \times {3}^{2} + 5 \times 3 - 6) = 5 \\ {3}^{x} (36 + 15 - 6) = 5 \\ {3}^{x} \times 45 = 5 \\ {3}^{x} = \frac{1}{9} \\ x = - 2$

$5)$

замена:

${3}^{x} = t \\ {t}^{2} - 6t - 27 = 0 \\ d = 36 + 108 = 144 \\ t1 = (6 + 12) \div 2 = 9 \\ t2 = - 3 \\ \\ {3}^{x} = 9 \\ x = 2 \\ \\ {3}^{x} = - 3$

в последнем корней нет

Ответ: 2.

$6) {5}^{4(4 - 3x)} \geqslant {5}^{ - 3} \\ 16 - 12x \geqslant - 3 \\ - 12x \geqslant - 19 \\ x \leqslant \frac{19}{12}$

$7) {2}^{5} \times {2}^{4(5 - 3x)} < {2}^{ - 2(5x - 6)} \\ 5 + 4(5 - 3x) < - 2(5x - 6) \\ 5 + 20 - 12x < - 10x + 12 \\ - 2x < - 13 \\ x > 6.5$

$8) {4}^{x} (4 + 1) \geqslant 320 \\ {4}^{x} \geqslant 64 \\ {4}^{x} \geqslant {4}^{3} \\ x \geqslant 3$