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17 tháng 5 2017

Sửa đề:

\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

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

\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

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

\(=2001.\dfrac{1}{10}-3\)

\(=200,1-3=197,1\)

Vậy S = 197,1

17 tháng 5 2017

kcj

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)

$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$

$\Leftrightarrow 1=1+1+1+S$

$S=1-1-1-1=-2$

9 tháng 7 2017

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)

\(\Rightarrow2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)=\dfrac{2017}{10}\)

\(\Rightarrow\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{a+c}=201,7\)

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}=201,7\)

\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{a+c}{a+c}+\dfrac{b}{a+c}=201,7\)

\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)

\(\Rightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=201,7\)

\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=198,7\)

9 tháng 7 2017

Ta có: \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\)

\(=>2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2017}{10}\)

\(=>\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{c+a}=201,7\)

Mà 2017 = a+b+c nên ta có:

\(=>\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=201,7\)

\(=>1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)

\(=>\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=201,7-3=198,7\)

CHÚC BẠN HỌC TỐT....

Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)

\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)

Vậy A = 2013