The correct option is B 1 [ 1/1 + a^( n m) + 1/1 + a^( m n) = 1/( 1 + a^n/a^m) + 1/( 1 + a^m/a^n); = a^m/( a^m + a^n) + a^n/( a^m + a^n) = ( a^m + a^n)/ ...

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Byju's Answer

Standard IX

Mathematics

Exponential Representation for Irrational Numbers

1/1+an-m+1/1+...

Question

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

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a ^{m + n}

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a^{m - n}

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Solution

The correct option is B
1

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

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Similar questions

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

Q. Solve:

\frac{1}{1 + a^{m - n}} + \frac{1}{1 + a^{n - m}}

Simplify

\frac{1}{1 + a^{m - n}} + \frac{1}{1 + a^{n - m}}

Q. Simplify:

\frac{1}{1 + a^{n - m}} + \frac{1}{1 + a^{m - n}}

Q. Prove that

\frac{1}{m} log (1 + a^{m}) < \frac{1}{n} log (1 + a^{n})

, if

m > n

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11+a(n−m)+11+a(m−n)=?

The correct option is B 1 11+a(n−m)+11+a(m−n)=1(1+anam)+1(1+aman)=am(am+an)+an(am+an)=(am+an)(am+an)=1.

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

The correct option is B
1

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