Cho:
\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)
cho \(\dfrac{x_1}{x_2}=\dfrac{x_2}{x_3}=\dfrac{x_3}{x_4}...=\dfrac{x_{2016}}{x_{2017}}\)
chứng minh: \(\left(\dfrac{x_1+x_2+x_3+...+x_{2016}}{x_2+x_3+x_4+...+x_{2017}}\right)^{2016}=\dfrac{x_1}{x_{2017}}\)
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
\({x^2} _1+{x^2} _2+{x^2} _3+...+{x^2} _{2017} = {x_1+x_2+x_3+...+x_{2017} \over {2017}} \)
\(C/m: x_1=x_2=x_3=x...=x_{2017}\)
yêu cầu đề bài đâu mà chứng minh đc. Lầy :I
\({x^2} _1+{x^2} _2+{x^2} _3+...+{x^2} _{2017} = {x_1+x_2+x_3+...+x_{2017} \over {2017}}\)
\(C/m: x_1=x_2=x_3=x...=x_{2017}\)
\({x^2}_1+{x^2}_2+{x^2}_3+...+{x^2}_{2017} = {{x_1+x_2+x_3+...+x_{2017}}\over 2017}\)
\(C/m : x_1=x_2=x_3=x...=x_{2017}\)
dell hiểu sao nó cứ ra thế, vế phải là \(\frac{x_1+x_2+x_3+...+x_{2017}}{2017}\)\(\)
\({x^2}_1+{x^2}_2+{x^2}_3+...+{x^2}_{2017} = {x_1+x_2+x_3+...+x_{2017}\over 2017}\)
\(C/m: x_1=x_2=x_3=x...=x_{2017}\)
\({x^2}_1+{x^2}_2+{x^2}_3+...+{x^2}_{2017} = {x_1+x_2+x_3+...+x_{2017}\over 2017}\)
\(C/m: x_1=x_2=x_3=x...=x_{2017}\)
\(x^2_1+x^2_2+x^2_3+...+x^2_{2017}\)\(=\dfrac{\left(x_1+x_2+x_3+...+x_{2017}\right)^2}{2017}\)
\(Cm:x_1=x_2=x_3=...=x_{2017}\)
BĐT Cauchy-Schwarz:
\(\left(1+1+1+...+1\right)\left(x^2_1+x^2_2+...+x^2_{2017}\right)\ge\left(x_1+x_2+...+x_{2017}\right)^2\left(\text{2017 số 1}\right)\)
\(\Leftrightarrow2017\left(x^2_1+x^2_2+...+x^2_{2017}\right)\ge\left(x_1+x_2+...+x_{2017}\right)^2\)
\(\Leftrightarrow x^2_1+x^2_2+...+x^2_{2017}\ge\dfrac{\left(x_1+x_2+...+x_{2017}\right)^2}{2017}\)
Khi \(\dfrac{x_1}{1}=\dfrac{x_2}{1}=...=\dfrac{x_{2017}}{1}\Leftrightarrow x_1=x_2=...=x_{2017}\)
Bạn j j biết làm bài ơi, giải hộ với. Bạn chưa biết làm thì nghĩ hộ t với. Làm được tớ cho mấy cái kẹo mút này...
Ú hú hú. mai 2h là t die r, giúp cái đi!!! Meo~!
Cho \(\left(2017x_1-2016y_1\right)^2+\left(2017x_2-2016y_2\right)^2+...+\left(2017x_{2016}-2016y_{2017}\right)^2\le0\)
CMR: \(\frac{x_1+x_2+x_3+...+x_{2016}}{u+y_1+y_2+y_3+...+y_{2016}}=\frac{2016}{2017}\)
u ở mẫu là cái gì vậy ?
Chàng Trai 2_k_7
cho \(\frac{_{x_1}}{x_2}=\frac{x_2}{x_3}=\frac{x_3}{x_4}=\frac{x_4}{x_5}=...=\frac{x_{2008}}{x_{2009}}\). Chứng minh rằng: \(\left(\frac{x_1+x_2+x_3+x_4+...+x_{2008}}{x_2+x_3+x_4+x_5+...+x_{2009}}\right)^{2008}\) = \(\frac{x_1}{x_{2009}}\)