Đáp án:
$+)\quad M = -4\quad khi \quad a+b+c+d = 0$
$+)\quad M = 4\,\,\,\quad khi\quad a+b+c+d\ne 0$
Giải thích các bước giải:
$+)\quad a+b+c+d = 0$
$\to \begin{cases}a+b = - (c+d)\\b+c= -(d+a)\\c+d = -(a+b)\\d+a = -(b+c)\end{cases}$
$\to M = \dfrac{-(c+d)}{c+d}+\dfrac{-(d+a)}{d+a}+\dfrac{-(a+b)}{a+b}+\dfrac{-(b+c)}{b+c}$
$\to M = -1-1-1-1$
$\to M = -4$
$+)\quad a+b+c+d \ne 0$
Ta có:
$\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}$
$\to \dfrac{a}{b+c+d}+1=\dfrac{b}{c+d+a}+1=\dfrac{c}{d+a+b}+1=\dfrac{d}{a+b+c}+1$
$\to \dfrac{a+b+c+d}{b+c+d}=\dfrac{a+b+c+d}{c+d+a}=\dfrac{a+b+c+d}{d+a+b}=\dfrac{a+b+c+d}{a+b+c}$
$\to \dfrac{1}{b+c+d}=\dfrac{1}{c+d+a}=\dfrac{1}{d+a+b}=\dfrac{1}{a+b+c}$
$\to b+c+d = c +d + a = d+a+b = a+b+c$
$\to a = b = c = d$
Do đó:
$M =\dfrac{a+a}{a+a} +\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}$
$\to M = 1+1+1+1$
$\to M = 4$
