Question

Simplificati fracțiile:b)abc+bca+cab supra xyz+yzx+zxy și c)ab0ab supra 17017

Answer 1

abc+bca+cab=100a+10b+c+100b+10c+a+100c+10a+b=111(a+b+c)
xyz+yzx+zxy=100x+10y+z+100y+10z+x+100z+10x+y=111(x+y+z)
(abc+bca+cab)/(xyz+yzx+zxy)=[111(a+b+c)/111(x+y+z)]=
=(a+b+c)/(x+y+z)

aboab=10000a+1000b+10a+b=10010a+1001b=1001(10a+b)=1001xab (cu bara deasupra ab)
17017=1001x17
(ab0ab)/ 17017=(1001xab)/(1001x17)=ab/17