1) a=0,5 * 0,(6)+8/5:0,2=[tex] \frac{5}{10}* \frac{6}{9}+ \frac{8}{5}* \frac{10}{2}=\frac{1}{2}* \frac{2}{3}+ 8= \frac{1}{3} +8= \frac{25}{3} [/tex]
Inversul lui a=[tex] \frac{3}{25} [/tex]
2)Considerma S=30 lei
prima zi = 3/10*S=3*30/10=9 lei
a doua zi=4/7*rest , iar rest =S-prima zi=30-9=21
a doua zi =4*21/7=12 lei
iar a treia zi= S-prima zi- a doua zi=30-9-12=9 lei
3)Notam suma de bani cu S.
prima zi =S/3
a doua zi=1/3 * rest, rest = S-prima zi= S-S/3=2S/3, iar a doua zi =1/3 * 2S/3=2S/9
a treia zi= 1/3 * rest nou, rest nou=S-(prima zi + a doua zi)=S-(S/3+2S/9)=S-5S/9=4S/9, deci a treia zi =1/3*4S/9=4S/27
a patra zi=120 lei
Toate zilele adunate dau suma S, deci =>
S/3+2S/9+4S/27+120=S (aducem la acelasi numitor care este 27)
9S+6S+4S+3240=27S =>19S+3240=27S => 18S=3240 => S=180 ->suma pe care a avut-o Andrei.
