Question

Usuń niewymierność z mianowników.

a) 3/√2 ; (4√3)/(5√6) ; (4√3+2)/√7 ; (4-3√5)/√5

b) 3/∛2 ; (4 ∛4)/∛5 ; (5∛5+2)/∛4 ; (3-3∛3)/∛6

Proszę o rozpisywanie :)

np.1/√2=1/√2*√2/√2 =√2/√2^2 =√2/2

PS. 2^2 to jest 2 do potęgi 2

/ to kreska ułamkowa

Dam naj :)

Answer 1

a) 3/√2= 3/√2·√2/√2 = 3√2/2

   (4√3)/(5√6) =( 4√3/5√6) · (√6/√6) = (4√18)/(5·6) = (4√(9·2))/30= (4·3√2)/30=

                    = 12√2/30 = 2√2/5

   (4√3+2)/√7 = (4√3+2)/√7 · √7/√7= (4√21+2√7)/7

   (4-3√5)/√5 = (4-3√5)/√5  · √5/√5 = (4√5-√15)/5

b) 3/∛2 = (3·∛2·∛2)/(∛2·∛2·∛2) = (3∛4)/2

   (4∛4/∛5 =(4∛4·∛5·∛5)/(∛5·∛5·∛5) = (4∛100)/5

   (5∛5+2)/∛4 = (5∛5+2)·∛4·∛4/ (∛4·∛4·∛4) = (5∛5+2)·∛16 / 4= (5∛80+2∛16)/4=

                     = (5∛(8·10) + 2∛(8·2) )/4 = (10∛10 + 4∛2)/4= [2(5∛10+2∛2)]/4 =

                     = (5∛10+2∛2)/2

 (3-3∛3)/∛6 = [(3-3∛3)·∛6·∛6]/(∛6·∛6·∛6) = [(3-3∛3)·∛36]/6 =[3(1-∛3)·∛36]/6=

                   = [(1-∛3)·∛36]/2 = (∛36 - ∛108)/2 = (∛36 - ∛(27·4) )/2 =(∛36-3∛4)/2