Question

Nếu \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b}\) (với a,b\(\ne\)0; a\(\ne\)-b) thì giá trị của biểu thức \(\dfrac{b}{a}+\dfrac{a}{b}\) là:........

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b}\Leftrightarrow\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2=ab\)

\(\Rightarrow\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{a^2+b^2}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=\dfrac{-ab}{ab}=-1\)

Answer 2

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{a+b}{ab}\Rightarrow\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)

\(\Leftrightarrow ab=\left(a+b\right)^2\)

Vậy, \(\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{a^2+b^2}{ab}=\dfrac{a^2+2ab+b^2-2ab}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=\dfrac{-ab}{ab}=-1\)