đặt $\frac{2a}{3b}$= $\frac{3b}{4c}$ =$\frac{4c}{5d}$ =$\frac{5d}{2a}$= t
⇒$\frac{2a}{3b}$. $\frac{3b}{4c}$ .$\frac{4c}{5d}$ .$\frac{5d}{2a}$= $t^{4}$
⇔$\frac{t.t.t.t}{t.t.t.t}$= $t^{4}$
⇔1=$t^{4}$
⇒$t^{4}$ =±1
với t=1 (t>0)
⇒$\frac{2a}{3b}$+ $\frac{3b}{4c}$ +$\frac{4c}{5d}$ +$\frac{5d}{2a}$=4
với t=-1 (t<0)
⇒$\frac{2a}{3b}$+ $\frac{3b}{4c}$ +$\frac{4c}{5d}$ +$\frac{5d}{2a}$=-4
Vậy C=±4