Răspuns :
[tex]\it a,(b)+b,(a)= (a+b)\cdot1,(1)\ \ \ \ (*)\\\;\\ a,(b)+b,(a)= (a+b)+0,(b)+0,(a)=a+b+\dfrac{b}{9}+\dfrac{a}{9}= (a+b)+\dfrac{a+b}{9}=\\\;\\= (a+b)(1+\dfrac{1}{9})=(a+b)\cdot1\dfrac{10}{9} = (a+b)\cdot1,(1)[/tex]
[tex]\it a,(b)+b,(a)= (a+b)\cdot1,(1)\ \ \ \ (*)\\\;\\ a,(b)+b,(a)= (a+b)+0,(b)+0,(a)=a+b+\dfrac{b}{9}+\dfrac{a}{9}= (a+b)+\dfrac{a+b}{9}=\\\;\\= (a+b)(1+\dfrac{1}{9})=(a+b)\cdot1\dfrac{10}{9} = (a+b)\cdot1,(1)[/tex]