Solution 3.1:6c

From Förberedande kurs i matematik 1

The expression is so complicated that we first need to simplify it. We start with the three fractions, 13 , 15  and 12 , which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators

12−2113−15=1222−211333−1555=22−2133−55.

Then, multiply the top and bottom of the fraction by 2 so that we get rid of the fractions in the denominator

22−21233−552=222−22323−525=2−1323−525.

Now, we can multiply the top and bottom by the conjugate of the denominator 2+1 , to get an expression without roots in the denominator.

2−1323−525=2−1323−5252+12+1=(2)2−12323−525(2+1)=2−13232+3231−5252−5251=13232+323−5225−525=326+323−5210−525.