\(\frac{45.16-17}{45.15+28}=\frac{45.\left(15+1\right)-17}{45.15+28}=\frac{45.15+45-17}{45.15+28}=\frac{45.15+28}{45.15+28}=1\)
45 . ( 15 + 1) - 17 45 . 15 + 28
= _______________ = ____________ = 1
45 . 15 + 28 45 . 15 + 28
\(\frac{45.16-17}{45.15+28}=\frac{45.\left(15+1\right)-17}{45.15+28}=\frac{45.15+45.1-17}{45.15+28}\)\(=\frac{45.15+45-17}{45.15+28}=\frac{45.15+28}{45.15+28}=1\)
\(\frac{45.16-17}{45.15+28}\)
\(=\frac{45.\left(15+1\right)-17}{45.15+28}\)
\(=\frac{45.15+45-17}{45.15+28}\)
\(=\frac{45.15+28}{45.15+28}=1\)