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Lösung 2.1:1d

Aus Online Mathematik Brückenkurs 1

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Version vom 13:41, 22. Okt. 2008

After x3y2 are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator,

x3y2y1−1xy+1=x3y2y1−x3y21xy+x3y21=yx3y2−xyx3y2+x3y2=x3y−x2y+x3y2

where we have used

yx3y2xyx3y2=yx3yy=x3y=xyxxxyy=xxy=x2y.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:1d“