HOC24
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Ta có : \(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\)
\(\Rightarrow3+P=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\)
\(\Rightarrow3+P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a +b+c}{a+b}\)
\(\Rightarrow3+P=\left(a+b+c\right).\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\)
Mà \(a+b+c=2018;\) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2017}{2018}\) \(\left(a,b\in R\right)\)
\(\Rightarrow3+P=2018.\dfrac{2017}{2018}\)
\(\Rightarrow3+P=2017\)
\(\Rightarrow P=2014\)
Vậy \(P=2014\)
a,Ta có: \(390^5\div130^5=\left(3.130\right)^5\div130^5=3^5.130^5\div130^5=3^5=243\)
b, Ta có :\(120^3\div60 ^3=\left(2.60\right)^3\div60^3=2^3.60^3\div60^3=2^3=8\)
a, Ta có \(\left|x\right|=1,21\)
\(\Rightarrow\left[{}\begin{matrix}x=1,21\\x=-1,21\end{matrix}\right.\)
Vậy \(x\in\left\{1,21;-1,21\right\}\)