(2a-1)/(10a^2-a-2) б) (6a^2-5a+1)/1-4a^2)

a) $\frac{2a-1}{10a^{2}-a-2}=\frac{2a-1}{10(a-\frac{1}{2})(a+\frac{2}{5})}=\frac{2a-1}{5(2a-1)(a+\frac{2}{5})}=\frac{1}{5(a+\frac{2}{5})}=\frac{1}{5a+2}$

б) $\frac{6a^{2}-5a+1}{1-4a^{2}}=\frac{6(a-\frac{1}{2})(a-\frac{1}{3})}{(1-2a)(1+2a)}=\frac{3(2a-1)(a-\frac{1}{3})}{-(2a-1)(2a+1)}=\frac{3(a-\frac{1}{3})}{-(2a+1)}=\frac{3a-1}{-(2a+1)}$

$\frac{-(1-3a)}{-(2a+1)}=\frac{1-3a}{2a+1}$