Помогите со 2 заданием!

Решите задачу:

$\sqrt[3]{9^{2x+1}}= \frac{3}{ \sqrt[5]{3} }\\\\ \sqrt[3]{3^{2(2x+1)}}=3^1*3^{- \frac{1}{5}}\\\\ \sqrt[3]{3^{4x+2}}=3^{ \frac{4}{5}}\\\\3^{ \frac{4x+2}{3}}=3^{ \frac{4}{5}}\\\\ \frac{4x+2}{3}= \frac{4}{5}\\\\5(4x+2)=3*4\\\\4x+2= \frac{12}{5}\\\\4x=\frac{12}{5}-2\\\\4x=\frac{12}{5}-\frac{10}{5}\\\\4x=\frac{2}{5}\\\\x=\frac{2}{5}:4\\\\x=\frac{1}{10}\\\\x=0,1$