Question

Simplificati fractia abc+bca+cab supra 3a+3b+3c Ajutați-ma urgenttt

Answer 1

$\frac{abc+bca+cab}{3a+3b+3c}$

$\frac{100a+10b+c+100b+10c+a+100c+10a+b}{3(a+b+c)}$

$\frac{111a+111b+111c}{3(a+b+c)}$

$\frac{111(a+b+c)}{3(a+b+c)}$

$= \frac{111}{3} =37$

Answer 2

$\frac{abc+bca+cab}{3a+3b+3c}= \frac{(100a+10b+c+100b+10c+a+100c+10a+b)}{3(a+b+c)} = \frac{111a+111b+111c}{3(a+b+c)}$

$\frac{111(a+b+c)}{3(a+b+c)} = \frac{111}{3} =37$