Question

Ajutor! (sa fie ireductibile)

Answer 1

a) = (1 + 2 + 3 + 4) / 15 = 10/15 = 2/3

b) = 3/(4x25) + 1/(4x25) + 21/(4x25) = (3 + 1 + 21)/ 100 = 25/100 = 1/4

Answer 2

a )

1 / 15 + 2 / 15 + 3 / 15 + 4 / 15 =  2 / 3

=  ( 1 + 2 + 3 + 4 ) / 15 =

= 10 / 15 =

=  ( 10 : 5 ) / ( 15 : 5 ) =  

= 2 / 3

b )

4 x 25 = 2² x 5²

→ observ ca fractiile au acelasi numitor

→ aflu suma numaratorilor supra numitorul comun 4 x 25 = 100

3 / ( 4 x 25 ) + 1 / ( 2² x 5² ) + 21 / ( 2² x 25 ) = 1 / 4

= ( 3 + 1 + 21 ) / 100 =

= 25 / 100 =

= ( 25 : 25 ) / ( 100 : 25 ) =

= 1 / 4 → fractie ireductibila