Lösung 1.2:2d
Aus Online Mathematik Brückenkurs 1
If we divide up the denominators into their smallest possible integer factors,
\displaystyle \begin{align}
& 45=5\centerdot 9=5\centerdot 3\centerdot 3 \\
& 75=3\centerdot 25=3\centerdot 5\centerdot 5 \\
\end{align}
the expression can be written as
\displaystyle \frac{1}{5\centerdot 3\centerdot 3}+\frac{1}{3\centerdot 5\centerdot 5}
and then we see that the denominators have \displaystyle 3\centerdot 5 as a common factor. Therefore, if we multiply the top and bottom of the first fraction by \displaystyle 5 and the second by \displaystyle 3 , the result is the lowest possible denominator.
\displaystyle \begin{align}
& \frac{2}{5\centerdot 3\centerdot 3}\centerdot \frac{5}{5}+\frac{1}{3\centerdot 5\centerdot 5}\centerdot \frac{3}{3} \\
& \\
& =\frac{2}{5\centerdot 3\centerdot 3\centerdot 5}+\frac{3}{3\centerdot 5\centerdot 5\centerdot 3} \\
& \\
& =\frac{10}{225}+\frac{3}{225} \\
\end{align}
The lowest common denominator is \displaystyle 225 .