Cho a + b + c = 2010 và \(\frac{1}{a+b}\) + \(\frac{1}{b+c}\) + \(\frac{1}{c+a}\) = \(\frac{1}{3}\). Tính S = \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
a+b+c = 2010 => a+b=2010-c ; b+c=2010-a ; c+a=2010-b
=> S = a/2010-a + b/2010-b + c/2010-c = 2010/2010-a - 1 + 2010/2010-b -1 + 2010/2010-c - 1
= 2010/b+c - 1 + 2010/c+a - 1 + 2010/a+b - 1
= 2010.(1/b+c + 1/c+a + 1/a+b) - 3
= 2010.1/3 - 3 = 667
Vậy S = 667
Tk mk nha
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)
\(\Rightarrow S+3=\frac{2010}{3}\)
\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)
Ta có \(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
=\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{2010}{3}=670\)
\(\Rightarrow S=667\)
cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
\(\Rightarrow\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\left(a+b+c\right)=\left(a+b+c\right)\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{c+a}\right)=\frac{2010}{3}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2010}{3}-1-1-1\)
\(\Rightarrow S=667\)
cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
tính \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2010-\left(b+c\right)}{b+c}+\frac{2010-\left(c+a\right)}{c+a}+\frac{2010-\left(a+b\right)}{a+b}\)
\(=\frac{2010}{b+c}-\frac{b+c}{b+c}+\frac{2010}{a+b}-\frac{a+b}{a+b}+\frac{2010}{a+c}-\frac{a+c}{a+c}=\left(\frac{2010}{b+c}+\frac{2010}{a+b}+\frac{2010}{a+c}\right)-\left(1+1+1\right)\)
\(=2010.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)-3=2010.\frac{1}{3}-3=670-3=667\)
Cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Cho a + b + c = 100 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính giá trị biểu thức : \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Bài 1: Tìm số tự nhiên n để phân số \(\frac{7n-8}{2n-3}\)có GTLN.
Bài 2: Tìm x, biết: \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\).
Bài 3: Cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\).
Tính S=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Cho a+b+c=2014 và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2014}\).Tính S=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)+\left(1+\frac{c}{a+b}\right)\)
\(S+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(S+3=\frac{2014.1}{2014}=1\Rightarrow S=1-3=-2\)
cho a+b+c=2017 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=> (a+b+c).(1/a+b + 1/b+c +1/c+a) = 2017/90
=> a+b+c/a+b + a+b+c/b+c + a+b+c/c+a = 2017/90
=> 1 + c/a+b + 1 + a/b+c + 1 + b/c+a = 2017/90
=> a/b+c + b/c+a +c/a+b = 2017/90 - 3 = 1747/90
Vậy S = 1747/90
Tk mk nha