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Floricica20121
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Simplificati fractia abc+bca+cab supra 3a+3b+3c Ajutați-ma urgenttt

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HyperCoder
[tex] \frac{abc+bca+cab}{3a+3b+3c} [/tex]

[tex] \frac{100a+10b+c+100b+10c+a+100c+10a+b}{3(a+b+c)} [/tex]

[tex] \frac{111a+111b+111c}{3(a+b+c)} [/tex]

[tex] \frac{111(a+b+c)}{3(a+b+c)} [/tex]

[tex]= \frac{111}{3} =37[/tex]
Iakabcristina2
[tex] \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)} [/tex]

[tex] \frac{111(a+b+c)}{3(a+b+c)} = \frac{111}{3} =37[/tex]

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