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 \(M=\frac{1}{x_1^1}+\frac{1}{x_2^2}+\frac{1}{x_3^3}+...+\frac{1}{x_{2011}^{2011}}\)
Những câu hỏi liên quan
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=\(\frac{1}{x_1^1}+\frac{1}{x_2^2}+...+\frac{1}{x_{2011}^{2011}}\)
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=.....=\frac{x_{2010}-2010}{1}\)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 \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
cho 2011 số tự nhiên x1,x2,x3,....,x2011 thỏa mãn điều kiện
\(\frac{1}{^{x^{11}}_1}+\frac{1}{_2x^{11}}+.....+\frac{1}{_{2011}x^{11}}=\frac{2011}{2048}\) tính tổng
\(\frac{1}{_1x^1}+\frac{1}{_2x^2}+....+\frac{1}{_{2011}x^{2011}}\)
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 \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
\(\frac{x_1-1}{2010}=...=\frac{x_{2010}-2010}{1}=\frac{x_1+x_2+...+x_{2010}-\left(1+2+...+2010\right)}{2010+2009+...+1}\)
\(=\frac{2\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
Vậy thay vào ta được: \(x_1=x_2=...=x_{2010}=2011\)
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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
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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:
\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)
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
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