Question

Aratati ca fractia a10+b34+c56/d27+e32+f41 (barate,deci sunt numere de trei cifre)se poate simplifica prin 100

Answer 1

    
$\frac{\overline{a10}+\overline{b34}+\overline{c56}}{\overline{d27}+\overline{e32}+\overline{f41}} = \\ \\ =\frac{100a+10+100b+34+100c+56}{100d+27+ 100e+32 + 100f+41}= \\ \\ =\frac{100a+100b+100c+10+ 34+56}{100d+ 100e + 100f+27+32+41}= \\ \\ = \frac{100(a+b+c) +10+ 34+56}{100(d+ e + f)+27+32+41}= \\ \\ = \frac{100(a+b+c) +100}{100(d+ e + f)+100}= \\ \\ = \frac{100(a+b+c+1)}{100(d+ e + f+1)}= ~~~~~=\ \textgreater \ ~ \boxed{Se ~simplifica~cu~100} \\ \\ =\boxed{\frac{a+b+c+1}{d+ e + f+1}}$