Question

fie multimea M= {5,3 2/7,2 4/5,5/7,3/8, 1 1/3, 1 2/5, 7/23, 2 2/3, 3/4}
A) selectati toate perechile de numere reciproc inverse formate din elementele multimii M.
b) formatii multimea A, care contine toate perechile selectate si aflati card A.

Answer 1

Răspuns:

Explicație pas cu pas:

M= {5, 3 2/7, 2 4/5, 5/7 , 3/8 , 1 1/3, 1 2/5, 7/23, 2 2/3, 3/4} <=>

5 = 5/1  ; 3 2/7 = (3·7+2)/7 = 23/7 ; 2 4/5 = (2·5+4)/5 = 14/5

1 1/3 = (1·3+1)/3 = 4/3

1 2/5 = (1·5+2)/5 = 7/5 ; 2 2/3 = (2·3+2)/3 = 8/3  =>

M = {5/1 ; 23/7 ; 14/5 ; 5/7 ; 3/8 ; 4/3 ; 7/5 ; 7/23 ;  8/3 ; 3/4}

a) perechile inverse sunt :

{23/7 ; 7/23 } ; {5/7 ; 7/5} ; {4/3 ; 3/4} ; {3/8 ; 8/3}

b) A = {(23/7 ; 7/23) ; (5/7 ; 7/5) ; (4/3 ; 3/4} ; (3/8 ; 8/3)}

card A = 8 elemente

#copaceibrainly