Cho \(A=\left(x_1+2y_1\right)^2+\left(2x_2+4y_2\right)^2+.......+\left(100x_{100}+200y_{100}\right)\le0\)
Hỏi \(B=\frac{x_1+x_2+......+x_{100}}{y_1+y_2+......+y_{100}}=?\)
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 :
\(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-2y_{2015}\right)^{2016}\le0\)
Tính \(A=\frac{x_1+x_2+x_3+...+x_{2015}}{y_1+y_2+y_3+...+y_{2015}}\)
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 :
\(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-2y_{2015}\right)^{2016}\le0\)
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 \(\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+...+\left(2x_{2015}-3y_{2015}\right)\le0\)
\(\Leftrightarrow\left(2x_1-3y_1\right)^{2016}+\left(2x_2-3y_2\right)^{2016}+..+\left(2x_{2015}-3y_{2015}\right)^{2016}=0\)
\(\Leftrightarrow2x_1-3y_1=0;2x_2-3y_2=0;....;2x_{2015}-3y_{2015}=0\)
\(\Leftrightarrow2x_1=3y_1\)
\(2x_2=3y_2\)
............................
\(2x_{2015}=3y_{2015}\)
\(\Leftrightarrow2\left(x_1+x_2+...+x_{2015}\right)=3\left(y_1+y_2+...+y_{2015}\right)\)
\(\Leftrightarrow\)\(\frac{x_1+x_2+x_3+...+x_{2015}}{y_1+y_2+y_3+...+y_{2015}}=\frac{3}{2}\)
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
\(Cho\left(x_1\cdot a-y_1\cdot b\right)^{2n}+\left(x_2\cdot a-y_2\cdot b\right)^{2n}+\left(x_3\cdot a-y_3\cdot b\right)^{2n}+......+\left(x_m\cdot a-y_m\cdot b\right)^{2n}\le0\)
Với m,n ∈ N*
Chứng minh: \(\frac{x_1+x_2+x_3+.....+x_m}{y_1+y_2+y_3+.....+y_m}=\frac{b}{a}\)
Tìm \(x_1,x_2,...,x_{100}\) thỏa mãn
\(\sqrt{x_1^2-1^2}+2\sqrt{x_2^2-2^2}+...+100\sqrt{x_{100}^2-100^2}=\dfrac{1}{2}\left(x_1^2+x_2^2+...+x_{100}^2\right)\)
\(\sqrt{x_1^2-1^2}+2\sqrt{x^2_2-2^2}+...+100\sqrt{x_{100}^2-100^2}=\dfrac{1}{2}\left(x_1^2+x^2_2+...+x_{100}^2\right)\)
\(\Leftrightarrow2\sqrt{x_1^2-1^2}+4\sqrt{x^2_2-2^2}+...+200\sqrt{x_{100}^2-100^2}=x_1^2+x^2_2+...+x_{100}^2\)
\(\Leftrightarrow x_1^2-1-2\sqrt{x_1^2-1}+1+x^2_2-4-4\sqrt{x^2_2-4}+4+...+x^2_{100}-10000-200\sqrt{x_{100}^2-10000}+10000=0\)
\(\Leftrightarrow\left(\sqrt{x^2_1-1}-1\right)^2+\left(\sqrt{x^2_2-4}-2\right)^2+....+\left(\sqrt{x^2_{100}-10000}-100\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2_1-1}-1=0\\\sqrt{x^2_2-4}-2=0\\....\\\sqrt{x^2_{100}-10000}-100=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\sqrt{1^2+1}=\sqrt{2}\\x_2=\sqrt{2^2+4}=2\sqrt{2}\\....\\x_{100}=\sqrt{100^2+10000}=100\sqrt{2}\end{matrix}\right.\)