Question

Cho a;b;c;d\(\ne\)0

Biết:\(\dfrac{\text{b+c+d}}{\text{a}}\)=\(\dfrac{\text{c+d+a}}{\text{ }\text{b}}\)=\(\dfrac{\text{d+a+b}}{\text{c}}\)=\(\dfrac{\text{a+b+c}}{\text{d}}\)=\(k\)

Tính\(k\)

Answer 1

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

b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=3(a+b+c+d)/a+b+c+d=3

suy ra k=3

Answer 2

taco:\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)=>\(\dfrac{b+c+d}{a}+1=\dfrac{c+d+a}{b}+1=\dfrac{a+b+d}{c}+1=\dfrac{a+b+c}{d}+1=k+1\)=>\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}=k+1=\dfrac{a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=\dfrac{4.\left(a+b+c+d\right)}{a+b+c+d}=4\)

=>k+1=4

=>k=3