Question

решить уравнение $frac{(3x)^9*(9x^4)^3*x^2}{(3x^3)^5*(27x)^3} =-96$

Answer 1

$frac{(3x)^9*(9x^4)^3*x^2}{(3x^3)^5*(27x)^3} = -96 \ frac{3^9*x^9*3^6*x^{12}*x^2}{3^5*x^{15}*3^9*x^3} = -96 \ frac{3^{(9+6)}*x^{(9+12+2)}}{3^{(5+9)}x^{(15+3)}} = -96 \ frac{3^{15}*x^{23}}{3^{14}x^{18}} = -96 \ 3*x^{5}= -96 \ x^{5}=-32 \ x=2$