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![](https://rs.olm.vn/images/avt/0.png?1311)
sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{9}+\frac{1}{10}\)
\(=\left(\frac{1}{3}+\frac{1}{10}\right)+\left(\frac{1}{4}+\frac{1}{9}\right)+\left(\frac{1}{5}+\frac{1}{8}\right)+\left(\frac{1}{6}+\frac{1}{7}\right)\)
\(=\frac{13}{30}+\frac{13}{36}+\frac{13}{40}+\frac{13}{42}=\frac{13.\left(84+70+63+60\right)}{2520}=\frac{13.277}{2520}\)
Phân số \(\frac{13.277}{2520}\) tối giản nên \(a=13m\) \(\left(m\inℕ^∗\right)\)
Vậy \(a⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cm \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}< \frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\right)< \frac{1}{2}\)
\(\Leftrightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}< 2\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}< 1\)
ta cần chứng minh điều trên:
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}=1-\frac{1}{5}\)
\(\Leftrightarrow A< 1-\frac{1}{5}< 1\)
suy ra đpcm
Đặt biểu thức là A
Ta có:A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}\)
\(\Rightarrow\)A=\(\frac{1}{4}+\left(\frac{1}{16}+\frac{1}{36}\right)+\left(\frac{1}{64}+\frac{1}{100}\right)\)
\(\Rightarrow\)A<\(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< 1-\frac{1}{8}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{40}{50}=\frac{4}{5}\)
\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Từ đây ta suy ra
A > \(\frac{4}{5}+\frac{1}{2}+\frac{1}{100}=1,31>1\)
Câu hỏi của Thăng Phạm - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo bài bạn làm nhé!
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(A=\frac{1}{10}+\frac{99}{100}>1\)
=> A > 1