Question

Помогите пж срочнооо! ​

Answer 1

$(\displaystyle\frac{3c+1}{3c-1}-\frac{3c-1}{3c+1}):\frac{2c}{6c+2}=\frac{(3c+1)^2-(3c-1)^2}{(3c-1)(3c+1)}*\frac{6c+2}{2c}=\frac{2*6c}{(3c-1)(3c+1)}*\\ \\ \frac{2(3c+1)}{2c}=\frac{12c}{3c-1}*\frac{2}{2c}=\frac{6}{3c-1}*2=\frac{12}{3c-1}$