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HA
0
8 tháng 2 2016
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=a\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+b\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+c\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)=\(\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
\(=\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{a+c}+\frac{a}{a+c}\right)\)
=\(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+\left(1+1+1\right)=2010.\frac{1}{3}=670\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=667\)
VM
0
XO
18 tháng 10 2020
Ta có \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{bca}\)
Lại có\(\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c}{c}\)
=> \(\frac{b+c-a}{a}+2=\frac{a+c-b}{b}+2=\frac{a+b-c}{c}+2\)
=> \(\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Nếu a + b + c = 0
=> a + b = -c
=> b + c = -a
=> a + c = - b
Khi đó A = \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{bca}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)
Nếu a + b + c \(\ne\) 0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Khi đó A = \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a.2b.2c}{abc}=\frac{8abc}{abc}=8\)
Vậy khi a + b + c = 0 => A = -1
khi a + b + c \(\ne\)0 => A = 8
sửa đề:1+c/b chứ ko phải là a+c/b nhé bn
+)Xét a+b+c=0
=>a+b=-c;b+c=-a;c+a=-b
Khi đó \(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{c+a}{c}\right)\left(\frac{b+c}{b}\right)\)
\(=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=-1\)
+)Xét a+b+c \(\ne\) 0
Theo t/c dãy....:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)
=>a+b-c=c=>a+b=2c
b+c-a=a=>b+c=2a
a+c-b=b=>a+c=2b
\(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{2a.2b.2c}{a.b.c}=2.2.2=8\)
Vậy........................