\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=.....=\frac{x_{2010}-2010}{1}\)va \(x_1+x_2+x_3+...+x_{2011}=2\left(1+2+3+...+2011\right)\)
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 \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
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 \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
\(\frac{x_1-1}{2010}=...=\frac{x_{2010}-2010}{1}=\frac{x_1+x_2+...+x_{2010}-\left(1+2+...+2010\right)}{2010+2009+...+1}\)
\(=\frac{2\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
Vậy thay vào ta được: \(x_1=x_2=...=x_{2010}=2011\)
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 \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=...=\frac{x_{2010}-2010}{1}=\frac{\left(x_1-1\right)+\left(x_2-2\right)+...+\left(x_{2010}-2010\right)}{1+2+...+2010}\) (TC DTSBN)
\(=\frac{\left(x_1+x_2+...+x_{2010}\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=\frac{2.\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
\(\Rightarrow x_1-1=2010;x_2-1=2009;....;x_{2010}-2010=1\)
=> x1 = x2 = x3 =..... = x2010 = 2011
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: \(\left(\frac{x_1+x_2+x_3+x_4+...+x_{2008}}{x_2+x_3+x_4+x_5+...+x_{2009}}\right)^{2008}\) = \(\frac{x_1}{x_{2009}}\)
cho 2011 số tự nhiên thõa mãn điều kiện
\(\frac{1}{x_1^{11}}+\frac{1}{x_2^{11}}+\frac{1}{x_3^{11}}+...+\frac{1}{x_{2011}^{11}}=\frac{2011}{2048}\)
tính tổng \(M=\frac{1}{x_1^1}+\frac{1}{x_2^2}+\frac{1}{x_3^3}+...+\frac{1}{x_{2011}^{2011}}\)
Gọi i là đại diện cho các số từ 1 đến 2011
ĐKXĐ: \(a_i\ne0\left(i=1,2,3,..,2011\right)\)
Xét \(a_i=1\) Ta có: \(\frac{1}{a^{11}_i}=1>\frac{2011}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}>\frac{2011}{2048}\left(loai\right)\)
Xét \(a_i\ge2\) Ta có: \(\frac{1}{a^{11}_i}\le\frac{1}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}\le\frac{2011}{2048}\)
Dấu "=" xảy ra khi \(a_i=2\)
Thay vào ta có:
\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2M-M=\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)\)
\(\Rightarrow M=1-\frac{1}{2^{2011}}\)
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 2011 số tự nhiên x1;x2;...;x2011 thỏa mãn đk:
\(\frac{1}{x_1^{11}}+\frac{1}{x_2^{11}}+...+\frac{1}{x_{2011}^{11}}=\frac{2011}{2048}\) tính:
M=\(\frac{1}{x_1^1}+\frac{1}{x_2^2}+...+\frac{1}{x_{2011}^{2011}}\)
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:
\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)