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Question Number 18746 by khamizan833@yahoo.com last updated on 29/Jul/17

Answered by daffa22 last updated on 30/Jul/17

$$\frac{1}{(1+\frac{1}{q})+p}+\frac{1}{(1+\frac{1}{r})+q}+\frac{1}{(1+\frac{1}{p})+r}$$$$misal,$$$$\bullet (1+\frac{1}{q})=a$$$$\bullet (1+\frac{1}{r})=b$$$$\bullet (1+\frac{1}{p})=c$$$$makamenjadi,$$$$\frac{1}{a+p}+\frac{1}{b+q}+\frac{1}{c+r}$$$$\bullet a+p=1+\frac{1}{q}+p$$$$a+p=\frac{1}{q}(q+1)+\frac{1}{qr}$$$$a+p=\frac{1}{q}(1+q+\frac{1}{r})$$$$a+p=\frac{1}{q}(b+q)$$$$q(a+p)=b+q$$$$\bullet b+q=1+\frac{1}{r}+q$$$$b+q=\frac{1}{r}(r+1)+\frac{1}{pr}$$$$b+q=\frac{1}{r}(1+r+\frac{1}{p})$$$$b+q=\frac{1}{r}(c+r)$$$$r(b+q)=c+r$$$$\bullet c+r=1+\frac{1}{p}+r$$$$c+r=\frac{1}{p}(p+1)+\frac{1}{pq}$$$$c+r=\frac{1}{p}(1+p+\frac{1}{q})$$$$c+r=\frac{1}{p}(a+p)$$$$p(c+r)=a+p$$$$Jadi,$$$$\frac{1}{a+p}+\frac{1}{b+q}+\frac{1}{c+r}$$$$=\frac{1}{p(c+r)}+\frac{1}{q(a+p)}+\frac{1}{r(b+q)}$$$$=\frac{1}{p(c+r)}+\frac{1}{pq(c+r)}+\frac{1}{qr(a+p)}$$$$=\frac{1}{p(c+r)}+\frac{1}{pq(c+r)}+\frac{1}{pqr(c+r)}$$$$=\frac{r(q+1)+1}{pqr(c+r)}=\frac{qr+r+pqr}{c+r}$$$$=\frac{qr+r+pqr}{\frac{1}{p}(1+p+\frac{1}{q})}=\frac{pqr+pr+p.pqr}{1+p+\frac{1}{q}}$$$$=\frac{1+p+\frac{1}{q}}{1+p+\frac{1}{q}}=1$$

Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17

$$LHS=\frac{q}{1+q+pq}+\frac{r}{1+r+rq}+\frac{p}{1+p+pr}=$$$$=\frac{q}{pqr+q+pq}+\frac{rp}{p+rp+1}+\frac{p}{1+p+pr}=$$$$=\frac{1}{p+pr+1}+\frac{pr}{1+rp+1}+\frac{p}{1+p+pr}=$$$$=\frac{1+pr+p}{1+pr+p}=1.◼$$

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