Question 20909
-9squareroot5/squareroot3
square root of 3 can be rationalised by multiplying with square root of 3 .it is true for any number which requires rationalisation.so multiply n.r and d.r with that number that is all .here we get 
-(9squareroot5)*(squareroot3)/[(squareroot3)*(squareroot3)=-9square root(5*3)/3
=-(9squareroot15)/3=-3squareroot15