Nhân cả 2 vế của đẳng thức cho a + b + c ta được :
\(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{a+b+c}{90}\)
\(\Rightarrow a+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2007}{90}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2007}{90}-3=22,3-3=19,3\)
\(a+b+c=2007\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
tính f= \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\frac{1}{a+b}.\left(a+b+c\right)+\frac{1}{b+c}.\left(a+b+c\right)+\frac{1}{c+a}.\left(a+b+c\right)=\frac{1}{90}.\left(a+b+c\right)\)
\(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{a+b+c}{90}\)
\(1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2007}{90}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3=\frac{2007}{90}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2007}{90}-3\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{193}{10}\)
vậy \(f=\frac{193}{10}\)
Ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=\left(a+b+c\right).\frac{1}{90}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{a+b+c}{90}\)
\(\Rightarrow\frac{\left(a+b\right)+c}{a+b}+\frac{a+\left(b+c\right)}{b+c}+\frac{b+\left(a+c\right)}{c+a}=\frac{a+b+c}{90}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2007}{90}=\frac{223}{10}=22,3\)
\(\Rightarrow\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=22,3-3=19,3\)
Vậy \(f=19,3\)
S=a/b+c +b/c+a +c/a+b =2007-(b+c)/b+c + 2007-(c+a)/c+a +2007- (a+b)/a+b =(2007/b+c)-1+(2007/c+a)-1+(2007/a+b)-1 =2007.[(1/b+c)+(1/c+a)+(1/a+b)].3 =2007.1/90-3 =19,3
Đây là cách mình mới tìm ra.Tuy không cùng kết quả nhưng bạn cứ xem thử
ta có:f=\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
=\(\dfrac{2007-\left(b+c\right)}{b+c}+\dfrac{2007-\left(c+a\right)}{c+a}+\dfrac{2007-\left(a+b\right)}{a+b}\)
=\(\dfrac{2007-1}{1}+\dfrac{2007-1}{1}+\dfrac{2007-1}{1}\)
=2006+2006+2006
=4012+2006
=6018
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