Question

If l, m, n are positive integers such that Imn + In + mn + Im + 1 + m + n = 1000 then find l+m+n.​

Answer 1

Answer:

if (log x)/(b - c) = (log y)/(c - a) = (log z)/(a - b) then find x^b+c-a .y^c+a-b.z^a+b-c

Step-by-step explanation:

Please answer it

Answer 2

Answer:

Given that:

lmn=1

To prove:

1+l+m

−1

1

+

1+m+n

−1

1

+

1+n+l

−1

1

=1

Solution:

First term:

1+l+m

−1

1

=

1+l+

m

1

1

=

m+lm+1

m

Second term:

1+m+n

−1

1

=

1+m+

n

1

1

=

1+m+lm

1

(lmn=1∴

n

1

=lm)

Third term:

1+n+l

−1

1

=

1+n+l

−1

1

×

lm

lm

=

lm+lmn+l

−1

lm

lm

=

lm+1+m

lm

L.H.S. =

1+l+m

−1

1

+

1+m+n

−1

1

+

1+n+l

−1

1

=

m+lm+1

m

+

m+lm+1

1

+

m+lm+1

lm

=

m+lm+1

m+1+lm

=1

= R.H.S.