Chứng minh:
\(B=\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+....+\frac{9}{100!}
Những câu hỏi liên quan
Chứng minh rằng :
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...\frac{9}{100!}< \frac{1}{9!}\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Bạn tham khảo nhé
\(a)\)Đặt \(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}< 1\) ( đpcm )
Vậy \(A< 1\)
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Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}
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Chứng minh rằng \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\) < \(\frac{1}{9!}\)
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=\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9}\)
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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