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Lösung 1.2:5a

Aus Online Mathematik Brückenkurs 1

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Version vom 13:29, 22. Okt. 2008

We begin by calculating the numerator in the main fraction,

271−115=2715115−17157=271515−7=28715.

Note that we keep 715 as it is, and do not multiply it to give 105, because this will make the task of cancellation later simpler. We calculate the fraction on the right-hand side by multiplying top and bottom by 7158,

28715=2871587158715=82715.

If we now divide up 8 and 15 into their smallest possible integer factors 8=222 and 15=35, we see that the answer in simplified form will be

82715=2222735=22735=4105.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:5a“