Question

Witam.Proszę pomóżcie mi w rozwiązania tych czterech przykładów z matematyki.W zadaniach jest potęga o wykładniku wymiernym.Proszę a zarazem błagam o pomoc.Dam najlepszą odpowiedź za rozwiązanie tego zadania.

Answer 1

a)

= 3^(1/20) * 27^(1/12) = 3^(1/20) * (3^3)^(1/12) =

= 3^(1/20)* 3^(3/12) = 3^(1/20) * 3^(1/4) =

= 3^(1/20)* 3^(5/20) = 3^(1/20 + 5/20) = 3^(6/20) = 3^(3/10)

b)

= [ 3^(1/2) * 9^(1/3)*3 ^(-1/2) ]/[81^(1/2) * 27^(2/3) ] =

= [ 3^( 1/2 - 1/2) * (3^2)^(1/3)]/[9 * [ 27^(1/3)]^2 ] =

= [ 3^(0) * 3^(2/3)] /[ 9* 3^2] = [ 1* 3^(2/3)] /[ 3^2 * 3^2 ] =

= [ 3 ^(2/3)]/[ 3^4] = 3 ^[(2/3) - 4] = 3^( - 3  1/3) = 3^( - 10/3)

c)

= [ 5 -2)*(5 + 2)*( 5^(1/3) - 2 ^(1/3))]/[ 5*5^(1/3) + 2*5^(1/3) - 5*2^(1/3) -

- 2*2^(1/3) ] =

= [ 3*7*( 5^(1/3) - 3^(1/3))]/ [7* 5^(1/3) - 7*2^(1/3) ] =

= [ 3*( 5^(1/3) - 2^(1/3))]/ [ 5^(1/3) - 2^(1/3)] =   3

d)

=[ [ 2*2^(1/4)+ 3*2^(1/4) - 2*3^(1/4) - 3*3^(1/4) ]*[ 2^(1/4) + 3^(1/4)]] /

/ [ 2^(1/2) - 3 ^(1/2)] =

=[ [ 5*2^(1/4) - 5*3^(1/4)]*[ 2^(1/4) + 3^(1/4)]]/[ 2^(1/2) - 3^(1/2) ] =

= [ [ 5*( 2^(1/4) - 3^(1/4))*( 2 ^(1/4) + 3^(1/4))]/ [ 2^(1/2) - 3^(1/2)] =

= [ 5*( 2^(1/2) - 3^(1/2))]/[ 2^(1/2) - 3^(1/2)] = 5