Tìm \(x_1\)biết \(\frac{x_1-1}{9}=\frac{x_2-2}{8}=\frac{x_3-3}{7}=...=\frac{x_9-9}{1}\)và \(x_1+x_2+x_3+...+x_9=90\)
cho \(\frac{_{x_1}}{x_2}=\frac{x_2}{x_3}=\frac{x_3}{x_4}=\frac{x_4}{x_5}=...=\frac{x_{2008}}{x_{2009}}\). 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:
\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)
tìm \(x_1;x_2;x_3;......;x_{2011}\) biet
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=.....=\frac{x_{2010}-2010}{1}\)va \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=.....=\frac{x_{2010}-2010}{1}\)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 \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)
Tìm các số x1,x2,x3,x4,......x8,x9 biết:
a)\(\frac{x_1+1}{9}=\frac{x_2+2}{8}=\frac{x_3+3}{7}=.....=\frac{x_8+8}{2}=\frac{x_9+9}{1}\)và x1 +x2+...+x8+x9=90
b) \(\frac{x_1-1}{9}=\frac{x_2-2}{8}=\frac{x_3-3}{7}=...=\frac{x_8-8}{2}=\frac{x_9-9}{1}\)và và x1 +x2+...+x8+x9=900
(giải chi tiết hộ mình nha)
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 \(x_1+x_2+x_3+...+x_n=c\)
\(\left(a_1\ne0,a_2\ne0,....,a_n\ne0,a_1+a_2+....+a_n\ne0\right)\)