Question

If pqr=1,then solve 1÷(1+p+1/q)+1÷(1+q+1/r)+1÷(1+r+1/p)

Answer 1

✏ Understanding the concept:-

• Here denominator of first term is (pq+q+1).
• We obtain same denominator for second and third term.
• To do so, we have to put value of r from pqr = 1.
• it means

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf \:  \dfrac{1}{r} = pq \: and \: r \: = \dfrac{1}{pq }$

$\bf \:  ⟼ Let's \: do \: the \: problem \: now !!$

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☆ Consider

$\sf \:  \dfrac{1}{1 + p + \dfrac{1}{q} } + \dfrac{1}{1 + q + \dfrac{1}{r} } + \dfrac{1}{1 + r + \dfrac{1}{p} }$

$\sf \:  = \: \dfrac{q}{q + pq + 1} + \dfrac{1}{1 + q + pq } + \dfrac{1}{1 + \dfrac{1}{pq} + \dfrac{1}{p} }$

$\: \: \: \: \: \: \: \: \: \: \bf \:  [\because \: \dfrac{1}{r} = pq \: and \: r \: = \dfrac{1}{pq }]$

$\sf \:  = \: \dfrac{q}{q + pq + 1} + \dfrac{1}{1 + q + pq } + \dfrac{1}{\dfrac{pq + 1 + q }{pq } }$

$\sf \:  = \: \: \dfrac{q}{q + pq + 1} + \dfrac{1}{1 + q + pq } + \dfrac{pq }{pq + 1 + q }$

$\sf \:  = \: \dfrac{q + 1 + pq }{q + pq + 1}$

$\sf \:  = \: \dfrac{ \cancel{q + pq + 1}}{ \cancel{q + pq + 1}}$

$\sf \:  = \: 1$

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Answer 2

$</p><p>\frac{1}{ {1 + p + q}^{ - 1} } + \frac{1}{ {1 +q +r }^{ - 1} } + \frac{1}{ {1 + r + p}^{ - 1} } \\ \\ = \frac{1}{ {1 + p + q}^{ - 1} } + \frac{ {q}^{ - 1} }{ {q}^{ - 1} + 1 + {q}^{ - 1} {r}^{ - 1} } + \frac{p}{p + pr + 1} \\ \\ = \frac{1}{1 + p + {q}^{ - 1} } + \frac{ {q }^{ - 1} }{1 + {q}^{ - 1} + 1 } \\ \\ (∴pqr = 1⇒ {(qr)}^{ - 1} = p \:⇒ {q}^{ - 1} {r}^{ - 1} = p \: and \: pr = {q}^{ - 1}) \\ \\ = \frac{1 + {q}^{ - 1} + p }{1 + {q}^{ - 1} + p} = 1$