Question

Se știe ca a=11 și b+c=8.Calculați:
a)a.b+a.c=
b)2.a+3.b+3.c=
c)10.a+9.b+9.c=
d)a.b+a.c+25=
e)a.b+a.c-34=
f)7.b+7.c+9.a=
g)11.b+11.c-6.a=
h)3.a-3.b-3.c=
i)11.b+a.c=
VĂ ROG FRUMOS !!!

Answer 1

$a)ab+ac=a(b+c)=11*8=88 \\ b)2a+3b+3c=2*11+3(b+c)=22+3*8=22+24=46 \\ c)10a+9b+9b=10*11+9(b+c)=110+9*8=110+72=182 \\ d)ab+ac+25=a(b+c)+25=88+25=113 \\ e)ab+ac-34=1(b+c)-34=88-34=54 \\ f)7b+7c+9a=7*8+9*11=99+56=155 \\ g)11b+11c-6a=11(b+c)-6*11=11*8-66=88-66=22 \\ h)3a-3b-3c=3*11-3(b+c)=33-8*3=33-24=9 \\ i)11b+ac=ab+ac=a(b+c)=8*11=88$