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Daca 2ab + b3a + ab4 = 567, atunci aratati ca ab∴3. Acolo vine 3 puncte in linie.

Răspuns :

200+10a+b+100b+30+a+100a+10b+4=567
200+111a+111b+30+4=567
111a+111b+234=567
111a+111b=567-234
111(a+b)=333
a+b=333:111
a+b=3