Question

prove $\frac{2v+3}{4v+5}= \frac{1}{2}+ \frac{1}{2(4v+5)}$

Answer 1

r.h.s
=1/2+1/8v+10
=8v+12/16v+20
=2v+3/4v+5
that is prope

Answer 2

$\frac{2v+3}{4v+5} = \frac{1}{2}+\frac{1}{2(4v+5)}$
$\frac{2v+3}{4v+5} - \frac{1}{2(4v+5)}$
=$\frac{2(2v+3)}{2(4v+5)} - \frac{1}{2(4v+5)}$
=$\frac{2(2v+3)}{2(4v+5)} - \frac{1}{2(4v+5)}$
=$\frac{(4v+6)}{2(4v+5)} - \frac{1}{2(4v+5)}$
=$\frac{(4v+6-1)}{2(4v+5)}$
=$\frac{(4v+5)}{2(4v+5)}$
= 1/2