Question

3.80
se consideră numerele întregi: a=4, b= +3, c =-11. Calculaţi:
a) [(a - b) + c] + (a + c);
b) [(a - b)+(c - a)] + (b-c);
c) [(a + b)+c] + [(a - b)+c];
d) [(c - a) - b] - a.​

Answer 1

a)[(a-b)+c]+(a+c)=

=[(4-3)-11]+(4-11)=

=(1-11)+(-7)=

=-10-7=

=-17

b)[(a-b)+(c-a)]+(b-c)=

=[(4-3)+(-11-4)]+(3-4)=

=(1-15)-1=

=-14-1=

=-15

c) [(a+b)+c]+[(a-b)+c]=

=[(4+3)-11]+[(4-3)-11]=

=(7-11)+(1-11)=

=-6-10=

=-16

d)[(c-a)-b]-a=

=[(11-4)-3]-4=

=(7-3)-4=

=4-4=

=0