Question

a=equal to root 5 +1 up on root 5 minus 1 and b=root5-1 upon root 5+1 then find a2+ab+b2/a2-ab+b2

Answer 1

This we can do by rationalization of irrational numbers in the denominator.

$a=\frac{\sqrt5+1}{\sqrt5-1},\ \ b=\frac{\sqrt5-1}{\sqrt5+1}\\\\\frac{a}{b}=\frac{(\sqrt5+1)^2}{(\sqrt5-1)^2}=\frac{5+1+2\sqrt5}{5+1-2\sqrt5}\\\\Rationalizing,\ \ =\frac{(3+\sqrt5)(3+\sqrt5)}{(3-\sqrt5)(3+\sqrt5)}=\frac{9+5+6\sqrt5}{9-5}=\frac{7+3\sqrt5}{2}\\\\\frac{a^2+ab+b^2}{a^2-ab+b^2}=divide\ by\ b^2\ in\ Nr.\ and\ Dr.\\\\=\frac{\frac{a^2}{b^2}+\frac{a}{b}+1}{\frac{a^2}{b^2}-\frac{a}{b}+1}=\frac{(7+3\sqrt5)^2+2(7+3\sqrt5)+4}{(7+3\sqrt5)^2-2(7+3\sqrt5)+4}\\\\=\frac{112+48\sqrt5}{84+36\sqrt5}$

if you check the numerator and denominator, you see that it is 4/3.
answer is 4/3.