Question

Cho dãy tỉ số bằng nhau :\(\dfrac{2a+b+c+d}{a}\)=\(\dfrac{a+2b+c+d}{b}\)=\(\dfrac{a+b+2c+d}{c}\)=\(\dfrac{a+b+c+2d}{d}\)

Tính giá trị biểu thức : M= \(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)

Answer 1

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

\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d-a-2b-c-d}{a-b}=1\)

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

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

\(\Rightarrow M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}\)

\(\Rightarrow M=1+1+1+1\)

\(\Rightarrow M=4\)

Vậy \(M=4\)