[(a^-1)+(b^-2)] \ [(a^-2)+(b^-1)] simplifiy?

Question

[(a^-1)+(b^-2)] \ [(a^-2)+(b^-1)] simplifiy?

Answer 1

lets calculate separately, because it's a little complicated to understand te writing.

so first we have: [(a^-1)+(b^-2)]  = (1 / a) + (1 / b^2) =  b^2 / ab^2 + a/ab^ 2 = (b^2 +a) / ab^2.

second we have:

[(a^-2)+(b^-1)]  = (1/a^2) + (1/b) = (b/ ba^2) + (a^2 / ba^2) =  (a^2+b )/ ba^2 .

now, to divide an expression by another is the same as  to multiply with the reverse of the second expression.

in other  words: [(b^2 +a) / ab^2 ] / [(a^2+b )/ ba^2 ]   =

[(b^2 +a) / ab^2 ]*[ ba^2 / a^2+b] = [( b^2 +a) * ba^2 ] / [ab^2 * (a^2 +b)] = ( b^2 +a) * a / b(a^2 +b ) = (ab^2 +a^2) / a^2b +b^2.

Answer 2

[(a^-1) + (b^-2)] \ [(a^-2) + (b^-1)]

=[(1/a) + (1/(b^2))] \ [(1/(2^2)) + (1/b)]

=[ ((b^2) + a) / (a(b^2)) ) \ ( (b + (a^2)) / (a^2b)) ]

=[ ((b^2) + a)/(a(b^2)) ] x [ ((a^2)b)/(b + (a^2)) ]

=[ a(b^2) + (a^2) ] \ [ (b^2) + (a^2)b ]

=[a[(b^2) + a]] \ [b[b + (a^2)]]                              

Answer 3

(1/a + 1/b^2) / (1/a^2 + 1/b)

(b^2 + a)/ab^2 * a^2b /(b + a^2)

a(b^2 + a)/b(b+a^2)

Answer 4

I got (a^2 + b) / (a + b^2).

Mathematica simplified it further to get

(1 / a) + (a^2 / b^2) + (1 / b)

but I'm not really sure how.

Answer 5

what kind of math is this because im scared now

i bet i will have it soon