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\(\frac{a}{2b}\)=\(\frac{b}{2c}\) =\(\frac{c}{2d}\) =\(\frac{d}{2a}\)=\(\frac{a+b+c+d}{2a+2b+2c+2d}\)=\(\frac{a+b+c+d}{2\left(a+b+c+d\right)}\)=\(\frac{1}{2}\)
quên rùi............................
đáp số =2
Áp dụng TC DTSBN ta có :
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{a}{2b}=\frac{1}{2}\Rightarrow a=\frac{1}{2}.2b\Rightarrow a=b\) (1)
\(\Rightarrow\frac{b}{2c}=\frac{1}{2}\Rightarrow b=\frac{1}{2}.2c\Rightarrow b=c\) (2)
\(\Rightarrow\frac{c}{2a}=\frac{1}{2}\Rightarrow c=\frac{1}{2}.2a\Rightarrow c=a\) (3)
\(\Rightarrow\frac{d}{2a}=\frac{1}{2}\Rightarrow d=\frac{1}{2}.2a\Rightarrow d=a\) (4)
Từ (1);(2);(3):(4) \(\Rightarrow a=b=c=d\) .Thay vào A ta được :
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a+2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(=\frac{a}{2a}+\frac{4021a}{2a}+\frac{a}{2a}+\frac{a}{2a}=\frac{a+4021a+a+a}{2a}=\frac{4024a}{2a}=\frac{4024}{2}=2012\)
Vậy \(A=2012\)
Áp dụng dãy tỉ số bằng nhau ta có: \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{1}{2}\)
\(\Rightarrow a=b;b=c;c=d;d=a\) hay \(a=b=c=d\)
\(\Rightarrow A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d+\right)}=\frac{1}{2}\)=\(\frac{1}{2}\)
\(\Rightarrow2a=2b,2b=2c,2c=2d,2d=2a\)
\(\Leftrightarrow a=b=c=d\)
\(\Rightarrow A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}=\frac{2011d-2010a}{b+c}\)
\(\Leftrightarrow A=\frac{2011a-2010a}{a+a}+\frac{2011b-2010b}{b+b}+\frac{2011c-2010c}{c+c}+\frac{2011d-2010d}{d+d}\)
\(\Leftrightarrow A=\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}+\frac{d}{2d}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
Ta có : \(\frac{a}{2b}\) = \(\frac{b}{2c}\) = \(\frac{c}{2d}\) =\(\frac{d}{2a}\) =\(\frac{a+b+c+d}{2\left(a+b+c+d\right)}\) =\(\frac{1}{2}\) ( a,b,c,d>0)
\(\Rightarrow\) \(\frac{a}{2b}\) =\(\frac{1}{2}\) \(\Rightarrow\) a=b (1) \(\frac{c}{2d}\) =\(\frac{1}{2}\)\(\Rightarrow\)c=d (3)
\(\frac{b}{2c}\) = \(\frac{1}{2}\) \(\Rightarrow\) b=c (2) \(\frac{d}{2a}\)=\(\frac{1}{2}\) \(\Rightarrow\) d=a(4)
Từ (1) ,(2) ,(3) và (4) \(\Rightarrow\)a=b=c=d (5)
Từ (5) ta thấy :a=b ,a=c ,a=d
\(\Rightarrow\)\(\frac{2011a-200b}{c+d}\) + \(\frac{2011b-2010c}{a+d}\) +\(\frac{2011c-2010d}{a+b}\) + \(\frac{2011d-2010a}{b+c}\)
= \(\frac{2011a-2010b}{a+a}\) + \(\frac{2011a-2010a}{a+a}\) + \(\frac{2011a-2010a}{a+a}\) + \(\frac{2011a-2010a}{a+a}\)
= \(\frac{2011a-2010a+2011a-2010a+2011a-2010a+2011a-2010a}{2a}\)
= \(\frac{a+a+a+a}{2a}\)= \(\frac{4a}{2a}\)=2
KL : \(\frac{a}{2b}\) = \(\frac{b}{2c}\) = \(\frac{c}{2d}\) = \(\frac{d}{2a}\)(a,b,c,d>0) thì A =2