Các câu hỏi tương tự
\(\frac{9}{10}!+\frac{9}{11}!+.......+\frac{9}{1000}!< \frac{1}{9}\)
16 tháng 4 2019 lúc 19:43
CMR:
\(\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
1,Chứng minh rằng
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Chứng minh:
A=\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9}\)
Chứng minh rằng :
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+........+\frac{9}{1000!}<\frac{1}{9!}\)
CHỨNG MINH RẰNG:
\(\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
Chứng minh B <1/90
B =\(\frac{9}{10!}+\frac{10}{11!}+......+\frac{999}{1000!}\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)