Question

Помогите пожалуйста!!!!!

Answer 1

Ответ:

1.

A)

${0.3}^{x} = \frac{1000}{27} \\ { (\frac{3}{10} )}^{x} = \frac{ {10}^{3} }{ {3}^{3} } \\{ (\frac{3}{10} )}^{x} = {( \frac{10}{3}) }^{3} \\ { (\frac{3}{10} )}^{x} = { (\frac{3}{10} )}^{ - 3} \\ x = - 3$

Б)

$3 \times {( \frac{4}{9}) }^{x} + 7 \times { (\frac{2}{3}) }^{x} - 6 = 0$

Пусть (2/3)^x=y, тогда

$3 {y}^{2} + 7y - 6 = 0 \\ d = {b}^{2} - 4ac = {7}^{2} - 4 \times 3 \times ( - 6) = 49 + 72 = 121 \\ y1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 7 - \sqrt{121} }{2 \times 3} = \frac{ - 7 - 11}{6} = \frac{ - 18}{6} = - 3 \\ y2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 7 + \sqrt{121} }{2 \times 3} = \frac{ - 7 + 11}{6} = \frac{4}{6} = \frac{2}{3}$

${( \frac{2}{3}) }^{x} = - 3$

Нет решений

${( \frac{2}{3}) }^{x} = \frac{2}{3} \\ x = 1$

В)

${5}^{x + 2} - 12 \times {5}^{x - 1} = 565 \\ {5}^{x} \times {5}^{2} - 12 \times {5}^{x} \times {5}^{ - 1} = 565 \\ 25 \times {5}^{x} - \frac{12}{5} \times {5}^{x} = 565 \\ {5}^{x} (25 - \frac{12}{5} ) = 565 \\ {5}^{x} (24 \frac{5}{5} - 2 \frac{2}{5} ) = 565 \\ {5}^{x} \times 22 \frac{3}{5} = 565 \\ {5}^{x} \times \frac{113}{5} = 565 \\ {5}^{x} = 565 \times \frac{5}{113} \\ {5}^{x} = 25 \\ {5}^{x} = {5}^{2} \\ x = 2$

2.

A)

${3}^{2x - 4} \leqslant 27 \\ {3}^{2x - 4} \leqslant {3}^{3} \\ 2x - 4 \leqslant 3 \\ 2x \leqslant 3 + 4 \\ 2x \leqslant 7 \\ x \leqslant \frac{7}{2} \\ x \leqslant 3.5$

Б)

${7}^{ {x}^{2} - 5x} < { (\frac{1}{7}) }^{ 6} \\ {7}^{ {x}^{2} - 5x} < {7}^{ - 6} \\ {x}^{2} - 5x = - 6 \\ {x}^{2} - 5x + 6 = 0 \\ d = {b}^{2} - 4ac = {( - 5)}^{2} - 4 \times 1 \times 6 = 25 - 24 = 1 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - ( - 5) - \sqrt{1} }{2 \times 1} = \frac{5 - 1}{2} = \frac{4}{2} = 2 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - ( - 5) + \sqrt{1} }{2 \times 1} = \frac{5 + 1}{2} = \frac{6}{2} = 3$

2<x<3

x∈(2,3)