Từ \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b}\) (a,b \(\ne\)0)
<=> \(\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)
<=> \(\left(a+b\right)^2=ab\)
Ta có: \(\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{b^2+a^2}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=-\dfrac{ab}{ab}=-1\)