Ôn tập chương III : Thống kê

TH

1 tìm x biết

\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2018}+\frac{x-4}{2015}\)

TT
22 tháng 4 2019 lúc 6:03

Sửa đề: \(\frac{x-3}{2018}\rightarrow\frac{x-3}{2016}\)

\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)

\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)

\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-2019=0\) (Vì \(\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)\ne0\) )

\(\Leftrightarrow x=2019\)

Vậy \(S=\left\{2019\right\}\)

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