Math, asked by Vedant250204, 1 year ago

URGENT.....

First one will be rated BRAINLIST...

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Answered by shriyamlaa

1 / 1+a^(n-m)] + [1 / 1+a^(m-n)]

=[1/1+(a^n/a^m)]+[1/1+(a^m/a^n)]

=[1/(a^m+a^n/a^m)]+[1/(a^n+a^m/a^n)]

=[a^m/(a^m+a^n)]+[a^n/(a^n+a^m)]

=[a^m+a^n]/[a^m+a^n]

=1

shriyamlaa: i cant solve it in copy now but i can tell you theoritically

shriyamlaa: ->we know that a^(m-n)=a^m/a^n
->By using the above formula the problem can be solved easily
->(1/1+(a^n/a^m))+(1/1+(a^m/a^n))
->Take LCM of denominators of each terms individually
->(1/((a^m+a^n)/a^m))+(1/((a^n+a^m)/a^n))
->(a^m/(a^m+a^n))+(a^n/(a^n+a^m))
->By taking LCM, both numerator and denominator gets cancelled by 1.
->The answer is 1

Vedant250204: well...... why can't you?

shriyamlaa: you can do it with another way

shriyamlaa: suppose a=1,m=2,n=3
put these values in equation
solving it.
(1 / 1+1^(3-2)) + (1 / 1+1^(2-3)) =1

1

Vedant250204: because is very hard to understand this type of solution for me..

Vedant250204: in which class do u study?

Vedant250204: ……

shriyamlaa: 10th

shriyamlaa: assuming way method is simpelest method for this type of question

Answered by Anonymous

[1/1+(a^n/a^m)]+[1/1+(a^m/a^n)]

=[1/(a^m+a^n/a^m)]+[1/(a^n+a^m/a^n)]

=[a^m/(a^m+a^n)]+[a^n/(a^n+a^m)]

=[a^m+a^n]/[a^m+a^n]

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