if m²-n²=1, then [p∧m/p∧n]∧m+n=
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Answered by
2
Solution :
It is given that ,
m² - n² = 1 -----( 1 )
Now ,
( p^m/p^n )^m+n
= ( p^m-n )^m+n
= p^( m-n )( m + n )
= p^m²-n²
= p¹ [ from ( 1 ) ]
= p
••••
It is given that ,
m² - n² = 1 -----( 1 )
Now ,
( p^m/p^n )^m+n
= ( p^m-n )^m+n
= p^( m-n )( m + n )
= p^m²-n²
= p¹ [ from ( 1 ) ]
= p
••••
Answered by
2
(p^m/p^n)^m+n
=p^(m-n)^(m+n)
=p^{(m+n)*(m-n)}
=p^{m^2-n^2}
=p^{m²-n²}
p^1
=p
=p^(m-n)^(m+n)
=p^{(m+n)*(m-n)}
=p^{m^2-n^2}
=p^{m²-n²}
p^1
=p
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