Question

Cho \(\dfrac{a}{2b}\)=\(\dfrac{b}{2c}\)=\(\dfrac{c}{2d}\)=\(\dfrac{d}{2a}\)(a,b,c,d >0). Tính A=\(\dfrac{2011a-2010b}{c+d}\)+\(\dfrac{2011b-2010c}{a+d}\)+\(\dfrac{2011c-2010d}{a+b}\)+\(\dfrac{2011d-2010a}{b+c}\)=...

Answer 1

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

suy ra:\(a=b;b=c;c=d;d=a\)

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

\(A=\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}\)

\(A=\dfrac{c+c+c+c}{c+c}=2\)

vậy giá trị của A là 2