# Ôn tập toán 7

Biết $a+b+c=2010$$\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}$

Tính $\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{c}{c+a}$

Nguyễn Huy Tú 10 tháng 3 2017 lúc 19:59

Ta có: $\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\left(\dfrac{c}{a+b}+1\right)+\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)-3$

$=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}-3$

$=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3$

$=2010.\dfrac{1}{3}-3$

$=670-3$

$=667$

Vậy $\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=667$

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Nguyễn Thị Thảo 10 tháng 3 2017 lúc 19:21

Đặt $S=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{c}{c+a}$

$\Rightarrow S=\dfrac{2010-\left(a+b\right)}{a+b}+\dfrac{2010-\left(b+c\right)}{b+c}+\dfrac{2010-\left(c+a\right)}{c+a}$$\Rightarrow S=\dfrac{2010}{a+b}+\dfrac{2010}{b+c}+\dfrac{2010}{c+a}-3$

$\Rightarrow S=2010\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3$

$\Rightarrow S=2010.\dfrac{1}{3}-3$

$\Rightarrow S=670-3$

$\Rightarrow S=667$

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