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2014-10-27T18:54:14+02:00
A)a(b+c)=11·8=88
b)2ab+3(b+c)=2ab+3·8=2·11·b+24=22b+24, sau b(2a+3)+3c=b·(2·11+3)+3c=25b+3c daca le scadem obtinem 3b+3c+24=3(b+c)+24=3·8+24=24+24=48
c)10a+9b+9c=10·11+9(b+c)=110+9·8=110+72=182
d)a(b+c)+25=11·8+25=88+25=113
e)a(b+c)-34=11·8-34=88-34=54
f)7(b+c)+9·11=7·8+99=56+99=155
2 5 2