Question

cho \(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)=2 (1)  ;  \(\dfrac{1}{a^2}\)+\(\dfrac{1}{b^2}\)+\(\dfrac{1}{c^2}\)=2 (2)
c/m a+b+c= abc

 

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=2\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

\(\Rightarrow a+b+c=abc\)