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\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Ta có:
\(\frac{1}{51}>\frac{1}{75}\)
\(\frac{1}{52}>\frac{1}{75}\)
......................
\(\frac{1}{75}=\frac{1}{75}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25.\frac{1}{75}=\frac{1}{3}\)(1)
Ta có:
\(\frac{1}{76}>\frac{1}{100}\)
\(\frac{1}{77}>\frac{1}{100}\)
........................
\(\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25.\frac{1}{100}=\frac{1}{4}\)(2)
Từ (1) và (2) ta có:
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{7}{12}\)(5)
Ta có:
\(\frac{1}{51}=\frac{1}{51}\)
\(\frac{1}{52}< \frac{1}{51}\)
...................
\(\frac{1}{75}< \frac{1}{51}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=25.\frac{1}{51}< 25.\frac{1}{50}=\frac{1}{2}\)(3)
Ta có:
\(\frac{1}{76}=\frac{1}{76}\)
\(\frac{1}{77}< \frac{1}{76}\)
...................
\(\frac{1}{100}< \frac{1}{76}\)
\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}=25.\frac{1}{76}< 25.\frac{1}{75}=\frac{1}{3}\)(4)
Từ (3) và (4) ta có:
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}\)(6)
Từ (5) và (6)
\(\Rightarrow\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}< \frac{5}{6}\)
đpcm
Tham khảo nhé~
Ta có:\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};..........;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{2011.2012}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{2011}-\frac{1}{2012}\)
\(=1-\frac{1}{2012}< 1\)
ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};\frac{1}{5^2}< \frac{1}{4.5};...;\frac{1}{2011^2}< \frac{1}{2010.2011};\)\(\frac{1}{2012^2}< \frac{1}{2011.2012}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2010.2011}+\frac{1}{2011.2012}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)
\(=1-\frac{1}{2012}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\left(đpcm\right)\)
Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
5A=1/5=2/5^2+......+11/5^11
4A=1/5+1/5^2+......+1/5^11-11/5^12
20A=1+1/5+1/5^2+.....+1/5^10-11/5^11
16A=1-1/5^11+11/5^12-11/5^11
vi 1-1/5^11<1;11/5^12-11/5^11<0
16A<1
A<1/16
k cho minh nhe
bn tham khảo đây nhé :
Câu hỏi của Khanh Mai Lê - Toán lớp 6 - Học toán với OnlineMath
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