Question

Cho a/b=b/c=c/d=d/a và a+b+c+d khác 0. Tính giá trị của   A= \(\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}\)

Answer 1

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

=>a=b=c=d=>\(a+b=\frac{1}{2}\left(a+b+c+d\right)\)

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

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

vậy A=2

Answer 2

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

\(\Rightarrow a=b=c=d\)

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

\(\frac{2a-a}{a+a}.4=\frac{a}{2a}.4=\frac{4a}{2a}=2\)

vậy A=2