Cho day so \(x_1,x_2,x_3,...\) thoa man \(x_{n+1}=\frac{x_n-1}{x_n+1}\) voi n=1, 2, 3,...
Biet \(x_{2015}=\frac{-1}{3}\), Tinh gia tri cua B=\(x_{10}+x_3+x_{2016}\)
Tìm các số \(x_1,x_2,...,x_{n-1},x_n\), biết rằng:
\(\frac{x_1}{a_1}=\frac{x_2}{a_2}=\frac{x_3}{a_3}=....=\frac{x_{n-1}}{a_{n-1}}=\frac{x_n}{a_n}\)và \(x_1+x_2+x_3+...+x_n=c\)
\(\left(a_1\ne0,a_2\ne0,....,a_n\ne0,a_1+a_2+....+a_n\ne0\right)\)
Ta có:
\(\frac{x_1}{a_1}=\frac{x_2}{a_2}=...=\frac{x_n}{a_n}=\frac{x_1+x_2+...+x_n}{a_1+a_2+...+a_n}_n=\frac{c}{a_1+a_2+...+a_n}\)
\(\Rightarrow x_1=\frac{a_1.c}{a_1+a_2+...+a_n}\) các x còn lại tương tự
Cho n số thực \(x_1;x_2;x_3;...;x_n\left(n\ge3\right)\)
\(CMR:max\left\{x_1;x_2;x_3;...;x_n\right\}\ge\frac{x_1+x_2+...+x_n}{n}+\frac{\left|x_1-x_2\right|+\left|x_2-x_3\right|+...+\left|x_{n-1}-x_n\right|+\left|x_n-x_1\right|}{2n}\)
\(max\left\{x_1;x_2;...;x_n\right\}\ge\frac{x_1+x_2+...+x_n}{n}+\frac{\left|x_1-x_2\right|+\left|x_2-x_3\right|+...+\left|x_{n-1}-x_n\right|+\left|x_n-x_1\right|}{2n}\)
Đề Tuyển sinh lớp 10 chuyên toán ĐHSP Hà Nội 2012-2013
NGUỒN:CHÉP MẠNG,CHÉP Y CHANG CHỨ E KO HIỂU GÌ ĐÂU(vài dòng đầu)-lỡ như anh cần mak ko có key. ( VÔ TÌNH TRA TÀI LIỆU THÌ THẦY BÀI NÀY )
P/S:Xin đừng bốc phốt.
Để ý trong 2 số thực x,y bất kỳ luôn có
\(Min\left\{x;y\right\}\le x,y\le Max\left\{x,y\right\}\) và \(Max\left\{x;y\right\}=\frac{x+y+\left|x-y\right|}{2}\)
Ta có:
\(\frac{x_1+x_2+...+x_n}{n}+\frac{\left|x_1-x_2\right|+\left|x_2-x_3\right|+.....+\left|x_n-x_1\right|}{2n}\)
\(=\frac{x_1+x_2+\left|x_1-x_2\right|}{2n}+\frac{x_2+x_3+\left|x_2-x_3\right|}{2n}+.....+\frac{x_3+x_4+\left|x_3-x_4\right|}{2n}+\frac{x_4+x_5+\left|x_4-x_5\right|}{2n}\)
\(\le\frac{Max\left\{x_1;x_2\right\}+Max\left\{x_2;x_3\right\}+.....+Max\left\{x_n;x_1\right\}}{n}\)
\(\le Max\left\{x_1;x_2;x_3;.....;x_n\right\}^{đpcm}\)
Cho:
\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)
Chứng minh rằng với các số thực dương \(x_1,x_2,...,x_n\)ta có:
\(\frac{x_1}{x_2+x_n}+\frac{x_2}{x_3+x_1}+\frac{x_3}{x_2+x_4}+...+\frac{x_n}{x_{n-1}+x_1}\ge2,\forall n\ge4\).
P/s: chứng minh bằng quy nạp
Với \(n=4\) bđt \(\Leftrightarrow\)\(\frac{x_1}{x_4+x_2}+\frac{x_2}{x_1+x_3}+\frac{x_3}{x_2+x_4}+\frac{x_4}{x_3+x_1}\ge2\)
\(\Leftrightarrow\)\(\frac{x_1^2}{x_4x_1+x_1x_2}+\frac{x_2^2}{x_1x_2+x_2x_3}+\frac{x_3^2}{x_2x_3+x_3x_4}+\frac{x_4^2}{x_3x_4+x_4x_1}\ge2\) (1)
\(VT_{\left(1\right)}\ge\frac{\left(x_1+x_2+x_3+x_4\right)^2}{2\left(x_1x_2+x_2x_3+x_3x_4+x_4x_1\right)}\ge\frac{\left(x_1+x_2+x_3+x_4\right)^2}{2.\frac{\left(x_1+x_2+x_3+x_4\right)^2}{4}}=2\)
Giả sử bđt đúng đến n=k hay \(\frac{x_1}{x_k+x_2}+\frac{x_2}{x_1+x_3}+...+\frac{x_{k-1}}{x_{k-2}+x_k}+\frac{x_k}{x_{k-1}+x_1}\ge2\)
\(\Leftrightarrow\)\(\frac{x_2}{x_1+x_3}+...+\frac{x_{k-1}}{x_{k-2}+x_k}\ge2-\frac{x_1}{x_k+x_2}-\frac{x_k}{x_{k-1}+x_1}\)
Với n=k+1, cần cm \(\frac{x_1}{x_{k+1}+x_2}+\frac{x_2}{x_1+x_3}+...+\frac{x_{k-1}}{x_{k-2}+x_k}+\frac{x_k}{x_{k-1}+x_{k+1}}+\frac{x_{k+1}}{x_k+x_1}\ge2\)
hay \(\frac{x_1}{x_{k+1}+x_2}-\frac{x_1}{x_k+x_2}+\frac{x_k}{x_{k-1}+x_{k+1}}-\frac{x_k}{x_{k-1}+x_1}+\frac{x_{k+1}}{x_k+x_1}\ge0\) (2)
giả sử \(x_k=max\left\{a_1;a_2;...;a_{k+1}\right\}\)
\(VT_{\left(2\right)}=\frac{x_1\left(x_k-x_{k+1}\right)}{\left(x_k+x_2\right)\left(x_{k+1}+x_2\right)}+\frac{x_k\left(x_1-x_{k+1}\right)}{\left(x_{k-1}+x_1\right)\left(x_{k-1}+x_{k+1}\right)}+\frac{x_{k+1}}{x_k+x_1}>0\)
nhầm, chỗ giả sử là \(x_{k+1}=min\left\{x_1;x_2;...;x_{k+1}\right\}\)
tìm \(x_1,x_2,x_3.......,x_9\)
\(\frac{x_{1-1}}{9}=\frac{x_{2-2}}{8}=\frac{x_3-3}{7}=....=\frac{x_{9-9}}{1}\) và \(x_1+x_2+x_3+...+x_9=90\)
Cho n số:\(x_1;x_2;x_3;........;x_n\)với \(x_k=1\)hoặc -1 (k=1;2;3;.....;n)
CMR:\(x_1.x_2+x_2.x_3+......+x_{n-1}.x_n=0\)thì n chia hết cho 4
tìm \(x_1;x_2;x_3;......;x_{2011}\) biet
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=.....=\frac{x_{2010}-2010}{1}\)va \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
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
cho \(\frac{_{x_1}}{x_2}=\frac{x_2}{x_3}=\frac{x_3}{x_4}=\frac{x_4}{x_5}=...=\frac{x_{2008}}{x_{2009}}\). Chứng minh rằng: \(\left(\frac{x_1+x_2+x_3+x_4+...+x_{2008}}{x_2+x_3+x_4+x_5+...+x_{2009}}\right)^{2008}\) = \(\frac{x_1}{x_{2009}}\)