CMR:
\(\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
KN
Những câu hỏi liên quan
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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\(\frac{9}{10}!+\frac{9}{11}!+.......+\frac{9}{1000}!< \frac{1}{9}\)
1,Chứng minh rằng
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
Chứng minh rằng:\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Chứng minh:
A=\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9}\)
1) Cho \(A=\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}.CMR:A< \frac{1}{9!}\)
2) \(CMR:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ai giúp mk sẽ đc thưởng 3 tick , phải ghi chép đầy đủ nha
HÃY CHỨNG MINH :
\(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+......+\frac{999}{1000}< \frac{1}{9!}\)
\(\frac{9}{10!}+\frac{10}{11!}+...+\frac{999}{1000!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+...+\frac{1000-1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)
đpcm
Tham khảo nhé~
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Chứng minh rằng :
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+........+\frac{9}{1000!}<\frac{1}{9!}\)
Ta đặt biểu thức đã cho là A
suy ra A < (10-1)/10! + (11-1)/11! +...+ (1000-1)/1000!
=> A < 10/10! - 1/10! + 11/11! - 1/11! +...+ 1000/1000! - 1/1000!
=> A < 1/9! - 1/10! + 1/10! - 1/11! +...+ 1/999! - 1/1000!
=> A < 1/9! - 1/1000! < 1/9!
Vậy A < 1/9!
Chúc bạn hoc tốt
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CHỨNG MINH RẰNG:
\(\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
A = 9/10! + 9/11! + 9/12! +......+ 9/1000! < 9/10! + 10/11! + 11/12! +...+999/1000! = B
9/10! = 1/9! - 1/10!
10/11! = 1/10! - 1/11!
...
999/1000! = 1/999! - 1/1000!
=> B= 1/9! - 1/1000! < 1/9!
=> A < 1/9! \(\left(ĐPCM\right)\).
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