Đáp án:
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Giải thích các bước giải:
Xét $a+b+c=0$
$\to \begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}$
Ta có:
$M=(1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a})$
$=\dfrac{a+b}{b} . \dfrac{b+c}{c} . \dfrac{c+a}{a}$
$=\dfrac{-c}{b} . \dfrac{-a}{c} . \dfrac{c+a}{a}$
$=-1$
Xét $a+b+c\ne0$
Theo tính chất dãy tỉ số bằng nhau, ta có:
$\dfrac{a+b}{c}=\dfrac{c+a}{b}=\dfrac{b+c}{a}=\dfrac{a+b+c+a+b+c}{c+b+a}=\dfrac{2(a+b+c)}{a+b+c}=2$
$\to \begin{cases}\dfrac{a+b}{c}=2\\\dfrac{c+a}{b}=2\\\dfrac{b+c}{a}=2\end{cases}$
$\to \begin{cases}a+b=2c\\c+a=2b\\b+c=2a\end{cases}$
Ta có:
$M=(1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a})$
$=\dfrac{a+b}{b} . \dfrac{b+c}{c} . \dfrac{c+a}{a}$
$=\dfrac{(a+b)(b+c)(c+a)}{abc}$
$=\dfrac{2c . 2a . 2b}{abc}$
$=\dfrac{8abc}{abc}$
$=8$
Vậy $\left[\begin{array}{l}M=-1\\M=8\end{array}\right.$
