\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
K
Khách
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Những câu hỏi liên quan
30 tháng 6 2022
a: \(A=\left[6\cdot\dfrac{1}{27}+3\cdot\dfrac{1}{3}+1\right]:\dfrac{-4}{3}\)
\(=\left(\dfrac{2}{9}+2\right)\cdot\dfrac{-3}{4}\)
\(=\dfrac{20}{9}\cdot\dfrac{-3}{4}=\dfrac{-60}{36}=\dfrac{-5}{3}\)
b: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)
\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)
20 tháng 6 2017
e) \(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{68}}=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}{\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}=\dfrac{1}{3}:\dfrac{1}{4}=\dfrac{3}{4}\)
25 tháng 6 2021
Đặt \(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)
Do đó: \(A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)(1)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
Do đó: \(A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(2)
Từ (1) và (2) ta suy ra ĐPCM
21 tháng 6 2017
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
RM
21 tháng 6 2017
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
H
2
K
1 tháng 4 2018
Xét mẫu số : \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)(cộng 2 cái ngoặc đầu tiên và lấy 2 nhân với ngoặc thứ 3 thì đc kết quả như này)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{50}\)
=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
Vậy thay kết quả của mẫu vừa tính đc vào E, ta có :
\(E=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}}=\) \(\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}}=1\)
\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
= \(\dfrac{4}{3}\)