Lösung 1.2:6

Aus Online Mathematik Brückenkurs 1

When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:

23+21 2141−31 and 32−72

We can do this by multiplying the top and bottom of each fraction by 2 , 12 and 7 respectively, so as to get rid of the partial fractions:

23+21= 223+212=432+212=46+1=742141−31=211241−3112=6412−312=63−4=6−1=−632−72=372−727=2127−727=2114−2=1221

The whole expression therefore equals

74−621−1221

If we multiply the tops and bottoms of the fractions

47 , 12 and 2112 in the main fraction by their lowest common denominator, 712 , we obtain integers in the numerator and denominator:

74−621−1221=74−671221−1221712=76−217412−6712=6−2174−6712=−157−3812=1573812

By factorizing 12 , 15 and 38 ,

12=26=22315=3538=219

the answer can be simplified to

1573812=357219223=35153