Dao Trong Luan Coordinator
07/04/2018 at 05:03-
Alone 07/04/2018 at 08:27\(\left(2x_1-5y_1\right)^{2018}\ge0;\left(2x_2-5y_2\right)^{2018}\ge0;.......;\left(2x_{2019}-5y_{2019}\right)^{2018}\ge0\)
\(\Rightarrow\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+......+\left(2x_{2019}-5y_{2019}\right)^{2018}\ge0\)
Because \(\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+..........+\left(2x_{2019}-5y_{2019}\right)^{2018}\le0\)
\(\Rightarrow\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+......+\left(2x_{2019}-5y_{2019}\right)^{2018}=0\)
\(\Rightarrow2x_1=5y_1;2x_2=5y_2;..........;2x_{2019}=5y_{2019}\)
\(\Rightarrow\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=..........=\dfrac{x_{2019}}{y_{2019}}=\dfrac{5}{2}=\dfrac{x_1+x_2+.......+x_{2019}}{y_1+y_2+.......+y_{2019}}\)
Dao Trong Luan selected this answer. -
¤« 07/04/2018 at 14:38(2x1−5y1)2018≥0;(2x2−5y2)2018≥0;.......;(2x2019−5y2019)2018≥0
⇒(2x1−5y1)2018+(2x2−5y2)2018+......+(2x2019−5y2019)2018≥0
Because (2x1−5y1)2018+(2x2−5y2)2018+..........+(2x2019−5y2019)2018≤0
⇒(2x1−5y1)2018+(2x2−5y2)2018+......+(2x2019−5y2019)2018=0
⇒2x1=5y1;2x2=5y2;..........;2x2019=5y2019
⇒x1y1=x2y2=..........=x2019y2019=52=x1+x2+.......+x2019y1+y2+.......+y2019