Question

Cho\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)

tính P biết

\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)

Answer 1

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

( theo tính chất dãy tỉ số bằng nhau )

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\cdot2b\\b=\dfrac{1}{2}\cdot2c\\c=\dfrac{1}{2}\cdot2d\\d=\dfrac{1}{2}\cdot2a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\)

\(\Rightarrow P=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}=2\)