Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(x+1\right).\left(x+2\right).\left(x+3\right)...\left(x+2016\right)=2016\)
\(A=x\left(1+2+3+...+2016\right)=2016\)
\(A=x\cdot\frac{\left(2016+1\right).2016}{2}=x\cdot2033136=2016\)
\(\Rightarrow x=2016:2033136=\frac{2}{2017}\)
\(\Rightarrow\frac{2}{2017}< \frac{1}{2015}\)
\(\Rightarrow x< \frac{1}{2015}\)
Ta có :
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+\frac{x-4}{2014}=2^2\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)+\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-4}{2014}-1\right)=2^2-4\)
\(\Leftrightarrow\)\(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+\frac{x-2018}{2014}=4-4\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
Nên \(x-2018=0\)
\(\Rightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+\frac{x-4}{2014}=2^2\)
\(\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)+\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-4}{2014}-1\right)=0\)
\(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+\frac{x-2018}{2014}=0\)
\(\left(x-2018\right).(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014})=0\)
\(x-2018=0\left(Vì\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\right)\)
\(\Rightarrow x=2018\)
Ta có: \(\left\{{}\begin{matrix}x_1+x_2+x_3+...+x_{2016}+x_{2017}=0\\x_1+x_2=x_3+x_4=x_5+x_6=...=x_{2015}+x_{2016}=x_{2016}+x_{2017}=1\end{matrix}\right.\)
Từ \(x_1+x_2+x_3+...+x_{2016}+x_{2017}=0\)
\(\Rightarrow\left(x_1+x_2\right)+\left(x_3+x_4\right)+...+\left(x_{2015}+x_{2016}\right)+x_{2017}=0\)
\(\Rightarrow1+1+...+1+x_{2017}=0\)
\(\Rightarrow1008+x_{2017}=0\Leftrightarrow x_{2017}=-1008\)
Mà \(x_{2016}+x_{2017}=1\Leftrightarrow x_{2016}=1-x_{2017}=1009\)