Question

Rezolvați exercițiile 11,12,13,15

Answer 1

Răspuns:

11. inlocuiesti a= 1, 2, 3, 4, ...

pt a=3 S= 3-$\frac{3}{2} +\frac{3}{3} -\frac{3}{5} +\frac{3}{30}$= 3+1 - $\frac{3}{2} -\frac{3}{5} +\frac{1}{10}$=4 +$\frac{3}{2} - \frac{3}{5} + \frac{1}{10}$= $\frac{4*10 -3*5-3*2+1}{10}$=

S= $\frac{40-15-6+1}{10} = \frac{20}{10}$= 2   S= 2   a= 3

12. a) aducem la acelasi numitor

$\frac{4+2-1}{8}$= $\frac{5}{8}$    

b) $\frac{5- 7*3}{6}$ + [( $\frac{-10*3+ 11*5+2*6}{30}) +(\frac{-13*6+10*10+1* 15}{30} )]=$$=\frac{-16}{6} +\frac{-30+55+12}{30} + \frac{-78+100+15}{30} =$ $\frac{-16*5}{30}$ +$\frac{37}{30}$ + $\frac{37}{30}$= $\frac{-80+37+37}{30} = \frac{-6}{30} = \frac{-1}{5}$

c) $\frac{-1*6-1*3-1*2}{18} = \frac{-6-3-2}{18} =\frac{-11}{18}$

d) $\frac{19}{36}$ + [ ( $\frac{4}{9} -\frac{37}{24}$) - $\frac{5}{18}$] = $\frac{19}{36}$ + [($\frac{4*8- 37*3}{72}$) - $\frac{5}{18}$= $\frac{19}{36}$ - $\frac{79}{72}$ - $\frac{5}{18}$= $\frac{19*2-79-5*4 }{72} = \frac{38-79-20}{72}$=

= $\frac{-61}{72}$