Question

Cho a,b,c ≠0 thảo mãn a+b+c=\(\sqrt{\text{2019}}\);\(\dfrac{\text{1}}{\text{a}}\)+\(\dfrac{\text{1}}{\text{b}}\)+\(\dfrac{\text{1}}{\text{c}}\)=0

Tính A=\(a^2+b^2+c^2\)

 

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow ab+bc+ca=0\)

\(a+b+c=\sqrt{2019}\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2019\)

\(\Rightarrow a^2+b^2+c^2=2019\) ( vì \(ab+bc+ca=0\))

Answer 2

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\\ A=a^2+b^2+c^2\\ \Leftrightarrow A=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\\ \Leftrightarrow A=\left(\sqrt{2019}\right)^2-2\cdot0=2019\)