Cho $\frac{a}{b+c+d}$ = $\frac{b}{c+d+a}$ = $\frac{c}{d+a+b}$ = $\frac{d}{a+b+c}$  Tính giá trị biểu thức:  S= $\frac{a+b}{c+d}$ + $\frac{b+c}{d+a}$ +

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:

$M =\dfrac{a+a}{a+a} +\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}$

$\to M = 1+1+1+1$

$\to M = 4$