Cho (2x1-3y1)2016+(2x2-3y2)2016+............+(2x2015-3y2015)2016 nhỏ hơn hoặc bằng. Tính A=\(\frac{x_1+x_2+.......+x_{2015}}{y_1+y_2+.......+y_{2015}}\)
Những câu hỏi liên quan
Cho :
\(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-2y_{2015}\right)^{2016}\le0\)
Tính \(A=\frac{x_1+x_2+x_3+...+x_{2015}}{y_1+y_2+y_3+...+y_{2015}}\)
Cho :
\(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-2y_{2015}\right)^{2016}\le0\)
Tính \(A=\frac{x_1+x_2+x_3+...+x_{2015}}{y_1+y_2+y_3+...+y_{2015}}\)
Vì \(\left(2x_1-3y_1\right)^{2016}\ge0;\left(2x_2-3y_2\right)^2\ge0;......;\left(2x_{2015}-3y_{2015}\right)\ge0\)
nên \(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-3y_{2015}\right)\le0\)
\(\Leftrightarrow\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+..+\left(2x_{2015}-3y_{2015}\right)^{2016}=0\)
\(\Leftrightarrow2x_1-3y_1=0;2x_2-3y_2=0;....;2x_{2015}-3y_{2015}=0\)
\(\Leftrightarrow2x_1=3y_1\)
\(2x_2=3y_2\)
............................
\(2x_{2015}=3y_{2015}\)
\(\Leftrightarrow2\left(x_1+x_2+...+x_{2015}\right)=3\left(y_1+y_2+...+y_{2015}\right)\)
\(\Leftrightarrow\)\(\frac{x_1+x_2+x_3+...+x_{2015}}{y_1+y_2+y_3+...+y_{2015}}=\frac{3}{2}\)
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Bài 1:Cho \(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-3y_{2015}\right)^{2016}\le0\)
Tính A= \(\dfrac{x_1+x_2+...+x_{2015}}{y_1+y_2+...+y_{2015}}\)
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 \(\left(2x_1-3y_1\right)^{2004}+\left(2x_2-3y_2\right)^{2004}+...+\left(2x_{2015}-3y_{2015}\right)^{2004}\)
C/M rằng \(\frac{x_1+x_2+x_3...+x_{2004}+x_{2005}}{y_1+y_2+y_3+...+y_{2005}+y_{2005}}=\frac{3}{2}\)
Cho:
( 2017x1 - 2016y2 )2 + ( 2017x2 - 2016y2 )2 + ... + ( 2017x2016 - 2016x2016)2 \(\le\)0
Chứng minh \(\frac{x_1+x_2+...+x_{2016}}{y_1+y_2+...+y_{2016}}=\frac{2016}{2017}\)
( 2017x1 - 2016y2 )2 + ( 2017x2 - 2016y2 )2 + ... + ( 2017x2016 - 2016x2016)2
Chẳng có quy luật gì cả :)))
Hình như sai đề
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sửa đề: \(\left(2017x_1-2016y_1\right)^2+\left(2017x_2-2016y_2\right)^2+....+\left(2017x_{2016}-2016y_{2016}\right)^2\le0\)
\(\text{ Ta có: }VT\ge0\text{ mà }VP\le0\)
Dấu = xảy ra khi: \(\left(2017x_1-2016y_1\right)^2+\left(2017x_2-2016y_2\right)^2+....+\left(2017x_{2016}-2016y_{2016}\right)^2=0\)
\(\Rightarrow2017x_1-2016y_1=0;2017x_2-2016y_2=0;.....;2017x_{2016}-2016y_{2016}=0\)
\(\Rightarrow2017x_1=2016y_1;2017x_2=2016_{y2};.....\text{và }2017x_{2016}=2016y_{2016}\)
\(\Rightarrow x_1=\frac{2016y_1}{2017};x_2=\frac{2016y_2}{2017};....;x_{2016}=\frac{2016y_{2016}}{2017}\)
=> \(\frac{x_1+x_2+x_3+...+x_{2016}}{y_1+y_2+y_3+...+y_{2016}}=\frac{\frac{2016.\left(y_1+y_2+...+y_{2016}\right)}{2017}}{y_1+y_2+...+y_{2016}}=\frac{2016}{2017}\)(đpcm)
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Xem thêm câu trả lời
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 \(A=\left(x_1+2y_1\right)^2+\left(2x_2+4y_2\right)^2+.......+\left(100x_{100}+200y_{100}\right)\le0\)
Hỏi \(B=\frac{x_1+x_2+......+x_{100}}{y_1+y_2+......+y_{100}}=?\)
xét A \(\ge\) 0;có A\(\le\) 0=>A=0
từ đó tính được x;y thế vào B làm tiếp
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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
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