Question

cho a+b+c+d khác 0\(\frac{a}{b+c+d}\)=\(\frac{b}{a+c+d}\)=\(\frac{c}{a+b+d}\)=\(\frac{d}{a+b+c}\)

tính \(\frac{a+b}{c+d}\)=\(\frac{b+c}{a+d}\)=\(\frac{c+d}{a+b}\)=\(\frac{d+a}{b+c}\)

Answer 1

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\) =\(\frac{a+b+c+d}{b+c+d+a+c+d+a+b+d+a+b+c}\)

Vì a+b+c+d khác 0

=> b+c+d=a+c+d=a+b+d=a+b+c

=>a=b=c=d

Khi đó:

a + b = c+d

b+c= (a+d)

c+d=a+b

d+a=b+c

=>\(\frac{a+b}{c+d}=\frac{b+c}{a+d}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)

 

 

 

Answer 2

theo bài ra ta có:

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

=> \(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)

=> \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)

=> b+c+d = a+c+d => a=b

a+c+d = a+b+d => b = c

a+b+d = a+b+c => c = d

a+b+c = b+c+d => a = d

=> a = b = c = d

=> \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

vậy \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=4\)

Answer 3

cho mk bổ sung thêm :

th1: a+b+c+d = 0

=> a+b = -(c+d)

=> b+c = -(a+d)

=> a+d = -(b+c)

=> c+d = -(a+b)

=> \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=-1+-1+-1+-1=-4\)