Question

b.B=|x-2016|+|x-2017|+|x-2018|

Bài 3 Cho a+b+c=2028 và \(\dfrac{1}{a+b}\)+\(\dfrac{1}{b+c}\)+\(\dfrac{1}{c+a}=\dfrac{1}{3}\)

Tính giá trị của biểu thức Q=\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

Answer 1

b) Tìm min

\(SV=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\)

\(SV=\left|x-2016\right|+\left|2018-x\right|+\left|x-2017\right|\)

\(SV\ge\left|x-2016+2018-x\right|+\left|x-2017\right|=2+\left|x-2017\right|\ge2\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}2016\le x\le2018\\x=2017\end{matrix}\right.\Leftrightarrow x=2017\)

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

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=676\)

\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=676\)

\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=673\)

Answer 2

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