So sánh S và B, biết:
\(S=\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}\); \(B=\frac{9!}{1000}\)
Những câu hỏi liên quan
Cho \(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
So sánh S với 10
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
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So sánh :
A = \(\frac{9}{11^4}+\frac{5}{11^5}\) và B = \(\frac{5}{11^41}+\frac{9}{11^5}\)
Có: \(A-B=\frac{9}{11^4}+\frac{5}{11^5}-\left(\frac{5}{11^4}+\frac{9}{11^5}\right)\)
\(=9\left(\frac{1}{11^4}-\frac{1}{11^5}\right)-\left[5\left(\frac{1}{11^4}-\frac{1}{11^5}\right)\right]\)
\(=4\left(\frac{1}{11^4}-\frac{1}{11^5}\right)\)>0
Vậy A>B.
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\(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}};B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
P/S : So sánh A và B
A=-9/10^2011+-9/10^2011+-9/10^2010
B=-9/10^2011+-9/10^2010+-10^2010
So sanh 10^2011>10^2010
Suy ra A>B
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So sánh \(A=\frac{10^{2016}-1}{10^{2017}-11}\) và \(B=\frac{10^{2016}+1}{10^{2017}+9}\)
Ta có : \(A=\frac{10^{2016}-1}{10^{2017}-11}\)
\(\Leftrightarrow10.A=\frac{10.\left(10^{2016}-1\right)}{10^{2017}-11}=\frac{10^{2017}-10}{10^{2017}-11}\)
\(=\frac{10^{2017}-11+1}{10^{2017}-11}=1+\frac{1}{10^{2017}-11}\)
Lại có : \(B=\frac{10^{2016}+1}{10^{2017}+9}\)
\(\Leftrightarrow10.B=\frac{10\left(10^{2016}+1\right)}{10^{2017}+9}=\frac{10^{2017}+10}{10^{2017}+9}\)
\(=\frac{10^{2017}+9+1}{10^{2017}+9}=1+\frac{1}{10^{2017}+9}\)
Do : \(10^{2017}-11< 10^{2017}+9\) \(\Rightarrow\frac{1}{10^{2017}-11}>\frac{1}{10^{2017}+9}\)
\(\Rightarrow1+\frac{1}{10^{2017}-11}>1+\frac{1}{10^{2017}+9}\)
hay \(A>B\)
Vậy : \(A>B\)
So sánh các số sau:
a) \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}v\text{à}\left(\sqrt{1\frac{1}{9}}-\sqrt{\frac{9}{16}}\right):5\)
b) \(\sqrt{25+9}v\text{à}\sqrt{25}+\sqrt{9}\)
CMR: \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
Có: \(\frac{9}{10!}=\frac{9}{10!}\)
\(\frac{9}{11!}< \frac{10}{11!}=\frac{11-1}{11!}=\frac{11}{11!}-\frac{1}{11!}=\frac{1}{10!}-\frac{1}{11!}\)
\(\frac{9}{12!}< \frac{11}{12!}=\frac{12-1}{12!}=\frac{12}{12!}-\frac{1}{12!}=\frac{1}{11!}-\frac{1}{12!}\)
............
\(\frac{9}{1000!}< \frac{999}{1000!}=\frac{1000-1}{1000!}=\frac{1000}{1000!}-\frac{1}{1000!}=\frac{1}{999!}-\frac{1}{1000!}\)
\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{1}{1000!}< \frac{9}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{999!}-\frac{1}{1000!}\)
\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+...+\frac{1}{1000!}< \frac{10}{10!}-\frac{1}{1000!}=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)
\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
\(\Rightarrowđpcm\)
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đặt tên là B
B=910!+911!+912!+.............+91000!
Ta thấy :
910!=10−110!=19!−110!
911!<11−111!=110!−111!
91000!<1000−11000!=1999!−11000!
⇒B<19!−110!+110!−111!+............+1999!−11000!
B<19!−11000!
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\(\frac{9}{10}!+\frac{9}{11}!+.......+\frac{9}{1000}!< \frac{1}{9}\)
16 tháng 4 2019 lúc 19:43
CMR:
\(\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
\(\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}\)
\(=\frac{10-1}{10!}+\frac{11-2}{11!}+...+\frac{1000-991}{1000!}\)
\(=\frac{10}{10!}-\frac{1}{10!}+\frac{11}{11!}-\frac{1}{11!}+...+\frac{1000}{1000!}-\frac{1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\left(đpcm\right)\)
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1,Chứng minh rằng
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}< \frac{1}{9!}\)