Đáp án:
Giải thích các bước giải:
GT có:
$\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{a+c}\Rightarrow \frac{a+b}{ab}=\frac{c+b}{cb}=\frac{a+c}{ac}\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\\\left\{\begin{matrix}
& \frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c} & \\
& \frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a} & \\
& \frac{1}{a}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c} &
\end{matrix}\right.\Rightarrow \frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c$
Thế vào BT có:
$\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$
