Question

Tính

N=\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)

Biết

N=\(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)

GIẢI ĐƯỢC BẰNG HAI CÁCH THÌ CÀNG TỐT

Answer 1

TH1 : a+b+c+d=0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\a+d=-\left(b+c\right)\\b+c=-\left(a+d\right)\\c+d=-\left(a+b\right)\end{matrix}\right.\Rightarrow N=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(a+d\right)}{a+d}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}=-4\)

TH2 : a+b+c+d khác 0

Ta có :

\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\\ \Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

Trừ mỗi tỉ số cho một

=> a=b=c=d

=> M = 4