Question

Помогите решить уравнение!

Answer 1

По свойству:

$sqrt[n]{ {a}^{m} } = {a}^{ frac{m}{n} }$

${ ({4}^{x + 2} )}^{x} times sqrt[3]{ {32}^{x - 1} } = 64 \ \ {2}^{2x(x + 2)} times {2}^{ frac{5(x - 1)}{3} } = {2}^{6} \ \ {2}^{2 {x}^{2} + 4x + frac{5x - 5}{3} } = {2}^{6} \ \ 2 {x}^{2} + 4x + frac{5x - 5}{3} = 6 \ \ 6 {x}^{2} + 12x + 5x - 5 = 18 \ \ 6 {x}^{2} + 17x - 23 = 0 \ \ d = {17}^{2} - 4 times 6 times ( - 23) = 841 \ \ 1) : : : : x = frac{ - 17 - 29}{12} = frac{ - 46}{12} = - frac{23}{6} \ \ 2) : : : : x = frac{ - 17 + 29}{12} = - frac{12}{12} = - 1$

ОТВЕТ: - 23/6 ; - 1