Đáp án: $M=\dfrac18$
Giải thích các bước giải:
Ta có:
$\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}$
$\to \dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2(a+b+c)}{a+b+c}$
$\to \dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=2$
$\to \dfrac{2b+c-a}{a}+1=\dfrac{2c-b+a}{b}+1=\dfrac{2a+b-c}{c}+1=3$
$\to \dfrac{2b+c-a+a}{a}=\dfrac{2c-b+a+b}{b}=\dfrac{2a+b-c+c}{c}=3$
$\to \dfrac{2b+c}{a}=\dfrac{2c+a}{b}=\dfrac{2a+b}{c}=3$
$\to 2b+c=3a\to c=3a-2b$
Mà $\dfrac{2c+a}{b}=\dfrac{2a+b}{c}$
$\to \dfrac{2(3a-2b)+a}{b}=\dfrac{2a+b}{3a-2b}$
$\to \dfrac{6a-4b+a}{b}=\dfrac{2a+b}{3a-2b}$
$\to \dfrac{7a-4b}{b}=\dfrac{2a+b}{3a-2b}$
$\to (7a-4b)(3a-2b)=b(2a+b)$
$\to 21a^2-26ab+8b^2=2ab+b^2$
$\to 21a^2-28ab+7b^2=0$
$\to 3a^2-4ab+b^2=0$
$\to 3a^2-3ab-ab+b^2=0$
$\to 3a(a-b)-b(a-b)=0$
$\to (3a-b)(a-b)=0$
$\to a=b$ hoặc $3a-b=0\to b=3a$
Trường hợp $1: a=b\to c=3a-2b=a\to a=b=c$
$\to M=\dfrac{(3a-2a)(3a-2a)(3a-2a)}{(3a-a)(3a-a)(3a-a)}=\dfrac18$
Trường hợp $2: b=3a\to c=3a-2\cdot 3a=-3a<0\to $ loại
