Question

stiind că a+b+c=18 2a+b+2c=25 calculați (a+b)×(3a+2b+3c)​

Answer 1

Răspuns:

a=7

a=7b=11

a=7b=11c=0

a=7b=11c=0(a+b)×(3a+2b+3c)=774

Explicație pas cu pas:

a+b+c=18

2a+b+2c=25 => din ambele, prin scădere avem

2a+b+2c=25 - a+b+c=18 => a+c=7 => a=7-c

a+b+c=18 - a+c=7 => b=11

2a+b+2c=25 - b=11 => 2a+2c=14

2(7-c)+2c=14 => 14-2c+2c=14 => c=0

a+c=7 => a+0=7 => a=7

(a+b)×(3a+2b+3c)=

(7+11)×[(3×7)+(2×11)+(3×0)]=

18×(21+22+0)=

18×43=

774