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\(\frac{9}{10!}+\frac{10}{11!}+...+\frac{999}{1000!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+...+\frac{1000-1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)
đpcm
Tham khảo nhé~
Đặt \(S=\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+.....+\frac{1}{2014^2}\)
Ta có : \(S< \frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+.....+\frac{1}{2013.2014}\\\)
Đặt \(A=\frac{1}{9.10}+\frac{1}{10.11}+....+\frac{1}{2013.2014}\\ =>A=\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+......+\left(\frac{1}{2013}-\frac{1}{2014}\right)\\ =>A=\frac{1}{9}-\frac{1}{2014}\\ \)
Vậy A<\(\frac{1}{9}\)
Mà A>S =>S<\(\frac{1}{9}\)
Chứng minh rằng: \(\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}\)
Có: \(\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{1100}\right)^{2008}\)
\(\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}=\frac{9^{2008}}{11^{2008}\times\left(10^2\right)^{2008}}=\frac{9^{2008}}{11^{2008}\times100^{2008}}=\frac{9^{2008}}{\left(11\times100\right)^{2008}}=\frac{9^{2008}}{1100^{2008}}=\left(\frac{9}{1100}\right)^{2008}\)
Vì: \(\left(\frac{9}{1100}\right)^{2008}=\left(\frac{9}{1100}\right)^{2008}\Rightarrow\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}\)
Bài làm
\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Rightarrow\frac{22}{30}-\frac{27}{30}< x< \frac{11}{15}.\frac{10}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{11}{3}.\frac{2}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{22}{27}\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
Vậy \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
~ Chắc z ~
# Chúc bạn học tốt #
Ta có:\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Leftrightarrow\frac{110-135}{30}< x< \frac{11.10}{15.9}\)
\(\Leftrightarrow\frac{-15}{30}< x< \frac{22}{27}\)
(Vì x c Z)\(\Leftrightarrow-1< x< 1\Rightarrow x\in\left\{0\right\}\)
a )( 2/5+2/9-2/11)/(8/5+8/9-8/11)=2*(1/5+1/9-1/11)/8*(1/5+1/9-1/11)=2/8=1/4