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a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
A= 1-\(\frac{1}{2}\) +\(\frac{1}{3}\) - \(\frac{1}{4}\) +\(\frac{1}{5}\)- \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) - \(\frac{1}{100}\)
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\) - 2 ( \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{100}\) )
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\)
= \(\frac{1}{51}\) + \(\frac{1}{52}\) +...+ \(\frac{1}{100}\)
= (\(\frac{1}{51}\) + \(\frac{1}{52}\) + ... + \(\frac{1}{75}\) ) + ( \(\frac{1}{76}\) + \(\frac{1}{77}\) + ... + \(\frac{1}{100}\) )
Ta có : \(\frac{1}{51}\) > \(\frac{1}{52}\) > \(\frac{1}{53}\) > ... > \(\frac{1}{75}\)
\(\frac{1}{76}\) > \(\frac{1}{77}\) > \(\frac{1}{78}\) > ... > \(\frac{1}{100}\)
=> A > \(\frac{1}{75}.25\) + \(\frac{1}{100}.25\) = \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{7}{12}\)
=> A< \(\frac{1}{51}.25\) + \(\frac{1}{75}.25\) < \(\frac{1}{50}.25\) + \(\frac{1}{75}.25\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\)
Vậy \(\frac{7}{12}\) < A < \(\frac{5}{6}\)
Tick nha
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)
c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
Ta có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25\cdot\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)
\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25\cdot\frac{1}{100}=\frac{25}{100}=\frac{1}{4}\)
\(\Rightarrow A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\left(1\right)\)
Lại có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=25\cdot\frac{1}{50}=\frac{25}{50}=\frac{1}{2}\)
\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25\cdot\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)
\(\Rightarrow A< \frac{1}{2}+\frac{1}{3}=\frac{5}{6}\left(2\right)\)
Từ (1) và (2) => đpcm
\(\Rightarrow\)(1/1.2) + ( 1/ 3.4) + (1/.6) +...+(1/99.100)
\(\Rightarrow\)(\(\frac{1}{1}\)-1/2 +1/3 -1/4 +...+ 1/99 - 1/100)
\(\Rightarrow\)( 1 - 1/100)
\(=\)99/100
Ta có \(\frac{7}{12}\)=0,5833
\(\frac{99}{100}\)=0,99
\(\frac{5}{6}\)=0,8333
Vì 0,99 > 0,8333 > 0,58333
\(\)\(\Leftrightarrow\)\(\frac{99}{100}\)>\(\frac{5}{6}\)>\(\frac{7}{12}\)
Vậy A lớn nhất trong cả 3 số không phải như điều cần chứng minh.
Câu hỏi của Vũ Thị Kim Oanh - Toán lớp 7 - Học toán với OnlineMath
Tham khảo
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{60}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{50}+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
2/ \(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{7}{12}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}>\frac{7}{12}\)
Tương tự câu trên ta có: \(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(A=\frac{1}{51}+...+\frac{1}{60}+\frac{1}{61}+...+\frac{1}{70}+\frac{1}{71}+...+\frac{1}{80}+\frac{1}{81}+...+\frac{1}{90}+\frac{1}{91}+...+\frac{1}{100}\)
\(A< \frac{1}{50}+...+\frac{1}{50}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{70}+...+\frac{1}{70}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{90}+...+\frac{1}{90}\)
\(A< 10.\frac{1}{50}+10.\frac{1}{60}+10.\frac{1}{70}+10.\frac{1}{80}+10.\frac{1}{90}\)
\(A< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{5}{6}\)
Đặt B = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Đặt C = \(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{75.76}\)(sửa lại đề)
=> 151C = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{75.76}\)
=> 151C =\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
=> C = \(\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}{151}\)
Khi A = B : C
= \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{151}\right)=151\)
Vậy A = 151