Răspuns:
11. inlocuiesti a= 1, 2, 3, 4, ...
pt a=3 S= 3-[tex]\frac{3}{2} +\frac{3}{3} -\frac{3}{5} +\frac{3}{30}[/tex]= 3+1 - [tex]\frac{3}{2} -\frac{3}{5} +\frac{1}{10}[/tex]=4 +[tex]\frac{3}{2} - \frac{3}{5} + \frac{1}{10}[/tex]= [tex]\frac{4*10 -3*5-3*2+1}{10}[/tex]=
S= [tex]\frac{40-15-6+1}{10} = \frac{20}{10}[/tex]= 2 S= 2 a= 3
12. a) aducem la acelasi numitor
[tex]\frac{4+2-1}{8}[/tex]= [tex]\frac{5}{8}[/tex]
b) [tex]\frac{5- 7*3}{6}[/tex] + [( [tex]\frac{-10*3+ 11*5+2*6}{30}) +(\frac{-13*6+10*10+1* 15}{30} )]=[/tex][tex]=\frac{-16}{6} +\frac{-30+55+12}{30} + \frac{-78+100+15}{30} =[/tex] [tex]\frac{-16*5}{30}[/tex] +[tex]\frac{37}{30}[/tex] + [tex]\frac{37}{30}[/tex]= [tex]\frac{-80+37+37}{30} = \frac{-6}{30} = \frac{-1}{5}[/tex]
c) [tex]\frac{-1*6-1*3-1*2}{18} = \frac{-6-3-2}{18} =\frac{-11}{18}[/tex]
d) [tex]\frac{19}{36}[/tex] + [ ( [tex]\frac{4}{9} -\frac{37}{24}[/tex]) - [tex]\frac{5}{18}[/tex]] = [tex]\frac{19}{36}[/tex] + [([tex]\frac{4*8- 37*3}{72}[/tex]) - [tex]\frac{5}{18}[/tex]= [tex]\frac{19}{36}[/tex] - [tex]\frac{79}{72}[/tex] - [tex]\frac{5}{18}[/tex]= [tex]\frac{19*2-79-5*4 }{72} = \frac{38-79-20}{72}[/tex]=
= [tex]\frac{-61}{72}[/tex]
