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Lösung 2.1:5b

Aus Online Mathematik Brückenkurs 1

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Version vom 13:48, 22. Okt. 2008

We can factorize the denominators as

y2−2yy2−4=y(y−2)=(y−2)(y+2)[difference of two squares]

and then we see that the terms' lowest common denominator is y(y−2)(y+2) because it is the product that contains the smallest number of factors which contain both y(y−2) and (y−2)(y+2).

Now, we rewrite the fractions so that they have same denominators and start simplifying

1y2−2y−2y2−4=1y(y−2)y+2y+2−2(y−2)(y+2)yy=y+2y(y−2)(y+2)−2y(y−2)(y+2)y=y+2−2yy(y−2)(y+2)=−y+2y(y−2)(y+2).

The numerator can be rewritten as −y+2=−(y−2) and we can eliminate the common factor y−2,

−y+2y(y−2)(y+2)=−(y−2)y(y−2)(y+2)=−1y(y+2)=−1y(y+2).
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:5b“