Question

Cho dãy tỉ số bằng nhau : \(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
và a+b+c+d khác 0 thì giá trị của biểu thức M = \(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{a+d}{c+b}\)
= ?

Answer 1

Theo bài ra , ta có :

\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\\ \Rightarrow1+\dfrac{2a+b+c+d}{a}=1+\dfrac{a+2b+c+d}{b}=1+\dfrac{a+b+2c+d}{c}=1+\dfrac{a+b+c+2d}{d}\\ \Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

Ta có : a+b+c+d \(\ne\) 0 \(\Rightarrow\) a=b=c=d

Thay vào M :

\(\Rightarrow M=\dfrac{a+a}{a}=\dfrac{b+b}{b}=\dfrac{c+c}{c}=\dfrac{d+d}{d}=4\)

Vậy M\(\in\) {4}

Answer 2

Cách khác:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}=\dfrac{2a+b+c+d-a-2b-c-d}{a-b}=1\)

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

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

\(=1+1+1+1=4\)

Vậy \(M=4\)