Question

Cho \(a,b,c\in Z;abc\ne0,\frac{a^2+b^2}{2}=ab;\frac{b^2+c^2}{2}=bc,\frac{a^2+c^2}{2}=ac\)

Tính : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).

Answer 1

\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+\frac{a}{b}+\frac{b}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

\(=\left(1+1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

\(=3+\frac{a^2+b^2}{ab}+\frac{a^2+c^2}{ac}+\frac{b^2+c^2}{bc}\)

\(=3+\frac{a^2+b^2}{\frac{a^2+b^2}{2}}+\frac{a^2+c^2}{\frac{a^2+c^2}{2}}+\frac{b^2+c^2}{\frac{b^2+c^2}{2}}\)

\(=3+2+2+2=9\)