3. Zadatak

Prvo rješenje.

Traženi izraz jednak je

$$\left(\frac1{ab}-\frac1{cd}\right)\left(\frac1c+\frac1d\right) =\frac{cd-ab}{abcd}\cdot\frac{d+c}{cd} =\frac{cd^2+c^2d-abd-abc}{abc^2d^2}.$$

Iz $\dfrac1a+\dfrac1b+\dfrac1c+\dfrac1d=0$ dobivamo $bcd+acd+abd+abc=0$,

odakle slijedi $bcd+acd=-abd-abc$.

Stoga je

$$\begin{aligned} \left(\frac1{ab}-\frac1{cd}\right)\left(\frac1c+\frac1d\right) &=\frac{cd^2+c^2d-abd-abc}{abc^2d^2}\\ &=\frac{cd^2+c^2d+bcd+acd}{abc^2d^2} =\frac{cd(d+c+b+a)}{abc^2d^2}\\ &=\frac{d+c+b+a}{abcd}. \end{aligned}$$

Kako iz $a+b+c+d+abcd=0$ slijedi $a+b+c+d=-abcd$,

dobivamo

$$\left(\frac1{ab}-\frac1{cd}\right)\left(\frac1c+\frac1d\right) =\frac{d+c+b+a}{abcd}=\frac{-abcd}{abcd}=-1.$$

Drugo rješenje.

Dijeljenjem jednakosti $a+b+c+d+abcd=0$ s $abcd$ dobivamo

$$\frac1{bcd}+\frac1{acd}+\frac1{abd}+\frac1{abc}+1=0,$$

a odatle

$$\frac1{cd}\left(\frac1b+\frac1a\right)+\frac1{ab}\left(\frac1d+\frac1c\right)=-1.$$

Kako je $\dfrac1b+\dfrac1a=-\dfrac1d-\dfrac1c$

imamo

$$\frac1{cd}\left(-\frac1d-\frac1c\right)+\frac1{ab}\left(\frac1d+\frac1c\right)=-1.$$

Konačno, izlučivanjem dobivamo

$$\left(\frac1c+\frac1d\right)\left(\frac1{ab}-\frac1{cd}\right)=-1.$$