Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{x_1}{x_2}=\frac{x_2}{x_3}=...=\frac{x_{2016}}{x_{2016} }=\frac{x_1+x_2+...+x_{2017}}{x_2+x_3+...+x_{2017}} \)( 2016 số)
\(=>\frac{x_1^{2016}}{x_2^{2016}}=\frac{x_2^{2016}}{ x_3^{2016}}=...=\frac{x_{2016}^{2016}}{x_{2017}^{2016}} =\frac{(x_1+x_2+...+x_{2016})^{2016}}{ (x_2+x_3+...+x_{2017})^{2016}}\)
Mà \(\frac{x_1^{2016}}{x_2^{2016}}=\frac{x_1}{x_2}. \frac{x_2}{x_3}.\frac{x_3}{x_4}...\frac{x_{2016}}{x_{2017}} =\frac{x_1}{x_{2017}}\)
=>đpcm
Nguyễn Tiến Dũng nói như z đứng đó k nhìn thấy làm sao mà làm đc bn ơi
\(\frac{x_1-1}{2010}=...=\frac{x_{2010}-2010}{1}=\frac{x_1+x_2+...+x_{2010}-\left(1+2+...+2010\right)}{2010+2009+...+1}\)
\(=\frac{2\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
Vậy thay vào ta được: \(x_1=x_2=...=x_{2010}=2011\)
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=...=\frac{x_{2010}-2010}{1}=\frac{\left(x_1-1\right)+\left(x_2-2\right)+...+\left(x_{2010}-2010\right)}{1+2+...+2010}\) (TC DTSBN)
\(=\frac{\left(x_1+x_2+...+x_{2010}\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=\frac{2.\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
\(\Rightarrow x_1-1=2010;x_2-1=2009;....;x_{2010}-2010=1\)
=> x1 = x2 = x3 =..... = x2010 = 2011
Ta có ; \(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{99-x_{99}}{1}\)
\(\Leftrightarrow\frac{1-x_1}{99}+1=\frac{2-x_2}{98}+1=\frac{3-x_3}{97}+1=...=\frac{99-x_{99}}{1}+1\)
\(\Leftrightarrow\frac{100-x_1}{99}=\frac{100-x_2}{98}=\frac{100-x_3}{97}=...=\frac{100-x_{99}}{1}\)
Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{100-x_1}{99}=\frac{100-x_2}{98}=\frac{100-x_3}{97}=...=\frac{100-x_{99}}{1}\)
\(=\frac{\left(100-x_1\right)+\left(100-x_2\right)+\left(100-x_3\right)=...=\left(100-x_{99}\right)}{1+2+3+...+98+99}\)
\(=\frac{100.99-\left(x_1+x_2+x_3+...+x_{99}\right)}{1+2+3+...+99}=\frac{100.99-4950}{\frac{99.100}{2}}=1\)
\(\Rightarrow x_i=100-\left(100-i\right)=i\)với \(i=1,2,3,...,99\)
\(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{99-x_{99}}{1}=\)\(\frac{\left(1+2+3+..+99\right)-\left(x_1+x_2+x_3+...+x_{99}\right)}{99+98+97+...+1}\)\(=\frac{4950-4950}{4950}=0\)
\(\Rightarrow1-x_1=2-x_2=3-x_3=...=99-x_{99}=0\)
\(\Rightarrow x_i=i-0\left(i=1,2,3,...,99\right)\)