Lösung 1.2:2d

Aus Online Mathematik Brückenkurs 1

Version vom 12:52, 11. Sep. 2008

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 .