Question

Буду благодарна.
Показатель-ые уравнения
(половина или меньше половины).

Answer 1

Ответ:

$1) {3}^{ {x}^{2} - 4x} = {3}^{0} \\ {x}^{2} - 4x = 0 \\ x1 = 0 \\ x2 = 4$

$2) {( \frac{1}{2} )}^{ {x}^{2} - x - 20} = {( \frac{1}{2}) }^{0} \\ {x}^{2} - x - 20 = 0 \\ d = 1 + 80 = 81 \\ x1 = (1 + 9) \times 2 = 5 \\ x2 = - 4$

$3) = {( \frac{1}{5}) }^{ {x}^{2} + x - 12} = 1 \\ {x}^{2} + x - 12 = 0 \\ d = 1 + 48 = 49 \\ x1 = 3 \\ x2 = - 4$

$4) \frac{x - 2}{x + 3} = 0 \\ x = 2$

$5) \frac{x + 3}{x - 4} = 0 \\ x = - 3$

$6) {7}^{x - 2} = {5}^{x - 2} \\ \frac{ {7}^{x - 2} }{ {5}^{x - 2} } = 1 \\ {( \frac{7}{5} )}^{x - 2} = 1 \\ x - 2 = 0 \\ x = 2$

$7) \frac{ {5}^{x - 4} }{ {10}^{x - 4} } = 1 \\ x - 4 = 0 \\x = 4$

$8) {2}^{2 {x}^{2} - 3x - 2} = 1 \\ 2 {x}^{2} - 3x - 2 = 0 \\ d = 9 + 16 = 25 \\ x1 = (3 + 5) \div 4 = 2 \\ x2 = - \frac{1}{2}$

$9) {3}^{ {x}^{2} - 2x } = {3}^{3} \\ {x}^{2} - 2x = 3 \\ {x}^{2} - 2x - 3 = 0 \\ d = 4 + 12 = 16 \\ x1 = 3 \\ x2 = - 1$

$10) {5}^{ {x}^{2} - x} = {5}^{2} \\ {x}^{2} - x = 2 \\ {x}^{2} - x - 2 = 0 \\ d = 1 + 8 = 9 \\ x1 = 2 \\ x2 = - 1$

$11) {5}^{ \frac{x}{x - 2} } = \frac{1}{25} \\ {5}^{ \frac{x}{x - 2} } = {5}^{ - 2} \\ \frac{x}{x - 2} = - 2 \\ x = - 2(x - 2) \\ x = - 2x + 4 \\ 3x = 4 \\ x = \frac{4}{3}$