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 \(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-3y_{2015}\right)\le0\)
\(\Leftrightarrow\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+..+\left(2x_{2015}-3y_{2015}\right)^{2016}=0\)
\(\Leftrightarrow2x_1-3y_1=0;2x_2-3y_2=0;....;2x_{2015}-3y_{2015}=0\)
\(\Leftrightarrow2x_1=3y_1\)
\(2x_2=3y_2\)
............................
\(2x_{2015}=3y_{2015}\)
\(\Leftrightarrow2\left(x_1+x_2+...+x_{2015}\right)=3\left(y_1+y_2+...+y_{2015}\right)\)
\(\Leftrightarrow\)\(\frac{x_1+x_2+x_3+...+x_{2015}}{y_1+y_2+y_3+...+y_{2015}}=\frac{3}{2}\)
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 \(\left(2x_1-3y_1\right)^{2004}+\left(2x_2-3y_2\right)^{2004}+...+\left(2x_{2015}-3y_{2015}\right)^{2004}\)
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 \(\left(2x_1-3y_1\right)^{2004}+\left(2x_2+3y_2\right)^{2004}+\left(2x_3+3y_3\right)^{2004}+...+\left(2x_{2005}+3y_{2005}\right)^{2004}\le0\)
Chứng minh rằng: \(\dfrac{x_1+x_2+x_3+...+x_{2005}}{y_1+y_2+y_3+...+y_{2005}}=1,5\)
Ta có \(\left\{{}\begin{matrix}\left(2x_1-3y_1\right)^{2004}\ge0\\......\\\left(2x_{2005}-3y_{2005}\right)^{2004}\ge0\end{matrix}\right.\) \(\forall x_1;x_2...x_{2005};y_1;y_2;...y_{2005}\)
Mà theo đề cho \(\left(2x_1-3y_1\right)^{2004}+...+\left(2x_{2005}-3y_{2005}\right)^{2004}\le0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x_1-3y_1\right)^{2004}=0\\\left(2x_2-3y_2\right)^{2004}=0\\.........\\\left(2x_{2005}-3y_{2005}\right)^{2004}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x_1-3y_1=0\\2x_2-3y_2=0\\........\\2x_{2005}-3y_{2005}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{3}{2}y_1\\x_2=\dfrac{3}{2}y_2\\.....\\x_{2005}=\dfrac{3}{2}y_{2005}\end{matrix}\right.\)
Từ đó ta có:
\(\dfrac{x_1+x_2+...+x_{2005}}{y_1+y_2+...+y_{2005}}=\dfrac{\dfrac{3}{2}y_1+\dfrac{3}{2}y_2+...+\dfrac{3}{2}y_{2005}}{y_1+y_2+...+y_{2005}}\)
\(=\dfrac{\dfrac{3}{2}\left(y_1+y_2+...+y_{2005}\right)}{y_1+y_2+...+y_{2005}}=\dfrac{3}{2}=1.5\) (đpcm)
Ghi lại đề đi bạn, nhìn qua dấu các biểu thức là biết bạn ghi sai đề rồi
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 \(\left(x_1+2y_1\right)^2+\left(2x_2+4y_2\right)^2+\left(3x_3+6y_3\right)^2+...+\left(100x_{100}+200y_{100}\right)^2\le0\)
Khi đó \(\frac{x_1+x_2+x_3+...+x_{100}}{y_1+y_2+y_3+...+y_{100}}=...\)
CÓ : (x1 + 2y1)^2 lớn hơn hoặc = 0 với mọi x,y
( 2x2 + 4y2)^2 lơn hơn hoặc = 0 với mọi x,y
.......
( 100x100 + 200y100 ) ^2 luôn lớn hơn hoặc = 0 với mọi x,y
=> (x1 + 2y1)^2 + (2x2+4y2)^2 +... + ( 100x100 + 200y100)^2 lớn hơn hoặc = 0 với mọi x,y
mà theo đề bài ta có ( x1 + 2y1)^2 + ( 2x2 + 4y2)^2 +.....+(100x100+200y100) ^2
nhỏ hơn hoặc =0
=> (x1+2y1)^2 + .........(100x100+ 200y100)^2=0
=>(x1+2y1)^2=0
tương tự đến(100x100 + 200y100)^2=0
từ đó bạn giải tiếp
Cho (2x1-3y1)2016+(2x2-3y2)2016+............+(2x2015-3y2015)2016 nhỏ hơn hoặc bằng. Tính A=\(\frac{x_1+x_2+.......+x_{2015}}{y_1+y_2+.......+y_{2015}}\)
\(Cho\left(x_1\cdot a-y_1\cdot b\right)^{2n}+\left(x_2\cdot a-y_2\cdot b\right)^{2n}+\left(x_3\cdot a-y_3\cdot b\right)^{2n}+......+\left(x_m\cdot a-y_m\cdot b\right)^{2n}\le0\)
Với m,n ∈ N*
Chứng minh: \(\frac{x_1+x_2+x_3+.....+x_m}{y_1+y_2+y_3+.....+y_m}=\frac{b}{a}\)
Cho PT \(2x^2-4mx+2m^2-1=0\). Tìm $m$ để PT có 2 nghiệm $x_1,x_2$ phân biệt thỏa:
\(\left(2x_1^{2016}-4mx_1^{2015}+\left(2m^2-1\right)x_1^{2014}+1\right)\left(2x_2^2+4mx_1+2m^2-9\right)< 0\)
\(\Delta'=4m^2-2\left(2m^2-1\right)=2>0\Rightarrow\) pt luôn có 2 nghiệm pb
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=\dfrac{2m^2-1}{2}\end{matrix}\right.\)
Do \(x_1\) là nghiệm nên:
\(2x_1^2-4mx_1+2m^2-1=0\Rightarrow x_1^{2014}\left(2x_1^2-4mx_1+2m^2-1\right)=0\)
Do \(x_2\) là nghiệm nên:
\(2x_2^2-4mx_2+2m^2-1=0\Rightarrow2x_2^2+2m^2-1=4mx_2\)
Bài toán trở thành:
\(\left(0+1\right)\left(4mx_2+4mx_1-8\right)< 0\)
\(\Leftrightarrow m\left(x_1+x_2\right)-2< 0\)
\(\Leftrightarrow2m^2-2< 0\)
\(\Leftrightarrow-1< m< 1\)