Lösung 3.1:6b

Aus Online Mathematik Brückenkurs 1

The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic

(3−2)2=(3)2−232+22=3−43+4=7−43.

Thus,

1(3−2)2−2=17−43−2=15−43

and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate 5+43 ,

15−43=15−435+435+43=5+4352−(43)2=5+4352−42(3)2=5+4325−163=−235+43=−235+43.