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a+b+c=4034=p
\(\frac{a}{c+b}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{p-\left(c+b\right)}{c+b}+\frac{p-\left(a+c\right)}{a+c}+\frac{p-\left(a+b\right)}{a+b}\)
\(=\frac{p}{c+b}+\frac{p}{a+c}+\frac{p}{a+b}-3=p\left[\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}\right]-3\)
\(=4034.\frac{1}{2}-3=2017-3=2014\)
a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)
\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{90}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2017}{90}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2017}{90}\)
\(\Rightarrow A+3=\frac{2017}{90}\)
\(\Rightarrow S=\frac{2017}{90}-3=\frac{1747}{90}\)
từ giả thiết, ta có
\(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}=\frac{1}{90}\)
Mà \(S=\frac{a}{2017-a}+\frac{b}{2017-b}+\frac{c}{2017-c}=-3+\frac{2017}{2017-a}+\frac{2017}{2017-b}+\frac{2017}{2017-c}\)
=-3+\(2017\left(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}\right)=-3+\frac{2017}{90}=\frac{1747}{90}\)
vậy ...
^_^
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
Ta có :
\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=2009.\frac{1}{7}=287\)
\(\Rightarrow S=287-3=284\)
Ok , mình sẽ làm !
Ta có :
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b}{c}-1+1=\frac{b+c}{a}-1+1=\frac{c+a}{b}-1+1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\left(1\right)\)
+) Trường hợp 1 : \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
Ta có :
\(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-a}{a}.\frac{-c}{c}.\frac{-b}{b}\)
\(\Leftrightarrow P=-1.\left(-1\right).\left(-1\right)=-1\)
+) Trường hợp 2 : \(a+b+c\ne0\)
Áp dụng tính chất của dãy tỉ số bằng nhau cho ( 1 ) , ta có :
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)
Ta lại có :
\(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)
\(\Leftrightarrow P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{c+b}{b}\)
\(\Leftrightarrow P=2.2.2=8\)
Vậy....................
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\frac{a}{b+c}=\frac{1}{2}\Rightarrow\frac{a}{b}=1\)Bn tự tính phần sau rồi thế vào đẳng thức đó mà tính
KQ: 8
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow S=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(\Rightarrow S=4034.\frac{1}{2}-3=2014\)
Vậy S=2014
key ban di chi gium n he khong k luon nhe