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nhuquynhdat

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0 \Longrightarrow ab+bc+ca=0$

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

$M=\dfrac{(a+b)(b+c)(c+a)}{abc}$

$M^2=\dfrac{[(a+b)(b+c)(c+a)]^2}{a^2b^2c^2}$

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

$\Longrightarrow M=1$
 
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