Question:
\[ \frac{a}{b}+\frac{b}{a}=1, \] then \[ a^2+b^2= \]
(a) 1
(b) -1
(c) 1/2
(d) 0
Solution:
\[ \frac{a}{b}+\frac{b}{a}=1 \]
\[ \frac{a^2+b^2}{ab}=1 \]
\[ a^2+b^2=ab \]
\[ \boxed{ab} \]
\[ \frac{a}{b}+\frac{b}{a}=1, \] then \[ a^2+b^2= \]
(a) 1
(b) -1
(c) 1/2
(d) 0
\[ \frac{a}{b}+\frac{b}{a}=1 \]
\[ \frac{a^2+b^2}{ab}=1 \]
\[ a^2+b^2=ab \]
\[ \boxed{ab} \]