Question

Daca a+b=18, calculați: a,(b)+b,(a) și a,(a)+b,(b)

Answer 1

 

a)

$\displaystyle\\\text{Transformam numere zecimale periodice in fractii.}\\\\a+b=18\\\\a,(b)+b,(a)=?\\\\a,(b)=\frac{\overline{ab}-a}{9}=\frac{10a+b-a}{9}=\frac{9a+b}{9} \\\\b,(a)=\frac{\overline{ba}-b}{9}=\frac{10b+a-b}{9}=\frac{9b+a}{9} \\\\a,(b)+b,(a)=\frac{9a+b}{9}+\frac{9b+a}{9}=\\\\=\frac{9a+b+9b+a}{9}=\frac{10a+10b}{9}=\frac{10(a+b)}{9}=\frac{10\times18}{9}=10\times2=\boxed{\bf20}$

b)

$\displaystyle\\a,(a)+b,(b)=?\\\text{Transformam numere zecimale periodice in fractii.}\\\\a,(a)=\frac{\overline{aa}-a}{9}=\frac{10a+a-a}{9}=\frac{10a}{9} \\\\b,(b)=\frac{\overline{bb}-b}{9}=\frac{10b+b-b}{9}=\frac{10b}{9}\\\\a,(a)+b,(b)=\frac{10a}{9}+\frac{10b}{9}=\\\\=\frac{10a+10b}{9}=\frac{10(a+b)}{9}=\frac{10\times18}{9}=10\times2=\boxed{\bf20}$