Ta có: \(b=a+1\Rightarrow b-a=1\)
\(\frac{1}{a}\times\frac{1}{b}=\frac{1}{a\times b}\)(1)
\(\frac{1}{a}-\frac{1}{b}=\frac{b-a}{a\times b}=\frac{1}{a\times b}\)(2)
Từ (1) và (2) suy ra \(\frac{1}{a}\times\frac{1}{b}=\frac{1}{a}-\frac{1}{b}\)
Ta có: b=a+1=>b-a=1
Theo bài ra ta có: \(\frac{1}{a}.\frac{1}{b}=\frac{1}{a.b}=\frac{b-a}{a.b}\left(b-a=1\right)=\frac{b}{a.b}=\frac{a}{a.b}=\frac{1}{a}-\frac{1}{b}\)
=>\(\frac{1}{a}.\frac{1}{b}=\frac{1}{a}-\frac{1}{b}\)