11+a(n−m)+11+a(m−n)=?11+a(n−m)+11+a(m−n)=?\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?" /> 11+a(n−m)+11+a(m−n)=?11+a(n−m)+11+a(m−n)=?\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?" /> 11+a(n−m)+11+a(m−n)=?11+a(n−m)+11+a(m−n)=?\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?" />

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$\frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} = ?$ $\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$

10
2¹/₂
31
4a^{m + n}

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} \\ = \frac{1}{(1 + \frac{a^{n}}{a^{m}})} + \frac{1}{(1 + \frac{a^{m}}{a^{n}})} \\ = \frac{a^{m}}{(a^{m} + a^{n})} + \frac{a^{n}}{(a^{m} + a^{n})} \\ = \frac{(a^{m} + a^{n})}{(a^{m} + a^{n})} \\ = 1 \end{aligned}

$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr}$

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", "text": "

\frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} = ?

$\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} \\ = \frac{1}{(1 + \frac{a^{n}}{a^{m}})} + \frac{1}{(1 + \frac{a^{m}}{a^{n}})} \\ = \frac{a^{m}}{(a^{m} + a^{n})} + \frac{a^{n}}{(a^{m} + a^{n})} \\ = \frac{(a^{m} + a^{n})}{(a^{m} + a^{n})} \\ = 1 \end{aligned}

$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{1}{1 + a^{(n - m)}} + \frac{1}{1 + a^{(m - n)}} \\ = \frac{1}{(1 + \frac{a^{n}}{a^{m}})} + \frac{1}{(1 + \frac{a^{m}}{a^{n}})} \\ = \frac{a^{m}}{(a^{m} + a^{n})} + \frac{a^{n}}{(a^{m} + a^{n})} \\ = \frac{(a^{m} + a^{n})}{(a^{m} + a^{n})} \\ = 1 \end{aligned}

$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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