Solve the rational equation $\frac{-1}{\sqrt[3]{x^2}}=0$

$\frac{u+2}{3u-1}+1=\frac{6-u}{u+1}$

$\frac{x+\frac{3}{6}}{1-\frac{3x}{6}}=\frac{2-x}{1+2x}$

$\frac{3x+1}{5x+7}=\frac{6x+11}{10x-3}$

$\frac{1}{\sqrt{\frac{\left(11\left(-1\right)\cdot x\right)^2}{2}+6x}}=0$

$\left(\sqrt{10}+3\right)^{\frac{x-3}{x-1}}=\left(\sqrt{10}-3\right)^{\frac{x+1}{x+3}}$

$\frac{x+1}{5}=\frac{7}{x-1}$

$\frac{x-3}{2}=\frac{6}{x-2}$