Question

[Nguyễn Thị Huệ]

cho \(\dfrac{x_1}{x_2}=\dfrac{x_2}{x_3}=\dfrac{x_3}{x_4}...=\dfrac{x_{2016}}{x_{2017}}\)

chứng minh: \(\left(\dfrac{x_1+x_2+x_3+...+x_{2016}}{x_2+x_3+x_4+...+x_{2017}}\right)^{2016}=\dfrac{x_1}{x_{2017}}\)

Answer 1

[Đạt Trần Tiến]

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