Question

2a-1/2a-2a/2a-1-1/2a-4a^2

Answer 1

$\frac{2a-1}{2a}- \frac{2a}{2a-1} - \frac{1}{2a-4 a^{2} } = \frac{2a-1}{2a} - \frac{2a}{2a-1} - \frac{1}{2a(1-2a)} = \frac{2a-1}{2a}- \frac{2a}{2a-1}+$$+ \frac{1}{2a(2a-1)}= \frac{(2a-1)(2a-1)-2a*2a+1}{2a(2a-1)}= \frac{4 a^{2}-4a+1-4 a^{2} +1 }{2a(2a-1)}= \frac{2-4a}{2a(2a-1)} =$$= \frac{2(1-2a)}{2a(2a-1)}=- \frac{1}{a}$