\(x^2_1+x^2_2+x^2_3+...+x^2_{2017}\)\(=\dfrac{\left(x_1+x_2+x_3+...+x_{2017}\right)^2}{2017}\)
\(Cm:x_1=x_2=x_3=...=x_{2017}\)
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 \(\frac{x_1+x_2+x_3+...+x_{2017}}{2017}\)\(\)
\({x^2}_1+{x^2}_2+{x^2}_3+...+{x^2}_{2017} = {x_1+x_2+x_3+...+x_{2017}\over 2017}\)
\(C/m: x_1=x_2=x_3=x...=x_{2017}\)
\({x^2}_1+{x^2}_2+{x^2}_3+...+{x^2}_{2017} = {x_1+x_2+x_3+...+x_{2017}\over 2017}\)
\(C/m: x_1=x_2=x_3=x...=x_{2017}\)
Cho:
\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)
Tìm GTNN của \(M=\frac{x^2_1+x^2_2+....+x_{2015}^2}{x_1\cdot\left(x_2+x_3+....+x_{2015}\right)}\)
mk nghĩ là dùng cô-si nhưng mãi ko ra
\(2014M=\frac{\left(x_1^2+2014x_2^2\right)+\left(x^2_1+2014x^2_3\right)+...+\left(x^2_1+x_{2015}^2\right)}{x_1\left(x_2+x_3+...+x_{2015}\right)}\)
\(2014M\ge\frac{2\sqrt{2014}x_1\left(x_2+x_3+...+x_{2015}\right)}{x_1\left(x_2+x_3+...+x_{2015}\right)}=2\sqrt{2014}\)
\(M\ge\frac{2}{\sqrt{2014}}\)
Dấu "=" xảy ra <=> \(\frac{\Leftrightarrow x_1}{\sqrt{2014}}=x_2=...=x_{2015}\)
Cho phương trình \(x^{2017}+ax^2+bx+c=0\) với các hệ số nguyên có 3 nghiệm \(x_1;x_2;x_3\). CMR nếu \(\left(x_1-x_2\right)\left(x_2-x_3\right)\left(x_3-x_1\right)\)không chia hết có 2017 thì \(a+b+c+1\)chia hết cho 2017