Question

va rog ajutati-ma e urgent

Answer 1

l)

{2+2 1/2 -2/3[0,(1)+0,(2)]}:3,1(6)=

2 si 1/2 se introduc intregii in fractie 2 si 1/2= (2×2+1)/2=5/2

{2+5/2 -2/3[0,(1)+0,(2)]}:3,1(6)=

0,(1)=1/9; 0,(2)=2/9

3,1(6)=(316-31)/9=285/9

2+5/2-2/3[1/9+2/9):285/9=

2+5/2-2/3(1/3×6/19)=

2+5/2-(2/3)×(1/3)×(6/19)=

([36+45-4)/18]×6/19=

[(81-4)/18]×6/19

=(77/18)×(6/19)

=77/(3×19)

=77/57

m)

8/35+[3/2-33/35]+(3/14-2/35)-(2-12/7+5/14)]}×3 si 1/3=

3 si 1/3 =se introduc intregii in fractie (3×3+1)/3=10/3

8/35+[(105-66)/(2×35)]+[(5×3-2×2)/{7×5×2)]-[(28-24+5)/14)]}×10/3=

8/35 +39/70+11/70-9/14)×(10/3)=

8/35+(5/7-9/14)×10/3=

8/35+[(10-9)/14]×10/3

=21/21

=1

Answer 2

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