so sanh cac phan so sau
a. \(\frac{999}{1000}\) va \(\frac{1000}{1001}\)
b.\(\frac{998}{1000}\) va \(\frac{1000}{1002}\)
c.\(\frac{1001}{1003}\) va \(\frac{1005}{1007}\)
ai tra loi duoc thi ket ban voi minh nhe
Những câu hỏi liên quan
so sanh 2 phan so sau:
\(\frac{2017}{1019}\)va \(\frac{2018}{2020}\)
ai tra loi duoc thi ket ban voi minh nhe
Ta có:
\(\frac{2017}{2019}=1-\frac{2}{2019}\)
\(\frac{2018}{2020}=1-\frac{2}{2020}\)
Vì \(\frac{2}{2019}>\frac{2}{2020}\)
=> \(1-\frac{2}{2019}>1-\frac{2}{2020}\)
=> \(\frac{2017}{2019}>\frac{2018}{2020}\)
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\(\frac{2017}{1019}>\frac{2018}{2020}\)
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Ta có: \(2017>1019\)
\(\Rightarrow\frac{2017}{1019}>1\)
\(\Rightarrow\frac{2017}{1019}>\frac{2017+1}{1019+1}=\frac{2018}{1020}>\frac{2018}{2020}\)
Vậy \(\frac{2017}{1019}>\frac{2018}{2020}\)
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tìm x biết
\(\frac{x-2}{1002}+\frac{x-4}{1001}+\frac{x-6}{1000}=\frac{x-8}{999}+\frac{x-10}{998}+\frac{x-12}{997}\)
Tính nhanh:\(\frac{\left(1+2\right)\times3}{\left(2+3\right)\times4}+\frac{\left(2+3\right)\times4}{\left(3+4\right)\times5}+...+\frac{\left(999+1000\right)\times1001}{\left(1000+1001\right)\times1002}+\frac{\left(1000+1001\right)\times1002}{\left(1001+1002\right)\times1003}\)
Tính nhanh:
\(987654321\cdot\frac{1+1\cdot2+2\cdot3+3\cdot4+4\cdot5+5\cdot6+6\cdot7+7\cdot8+8\cdot9+9\cdot...\cdot999+999\cdot1000+1000}{1000+1000\cdot1001+1001\cdot1002+1002\cdot1003+1003\cdot1004+1004\cdot1005+1005\cdot...\cdot9999+9999\cdot10000+10000}\)
SO SANH
A=\(\frac{3}{4}+\frac{8}{9}+............+\frac{999}{1000}\)voi 98 va 99
A = \(\frac{1001}{1000^2+1}+\frac{1001}{1000^2+2}+....+\frac{1001}{1000^2}+1000\)
A = \(\frac{1001}{1000^2+1}+\frac{1001}{1000^2+2}+....+\frac{1001}{1000^2}+1000\)
Chứng minh rằng 1 < A < 2 :
\(A=\frac{1001}{1000^2+1}+\frac{1001}{1000^2+2}+...+\frac{1001}{1000^2+1000}\)
$\frac{999}{1000}+\frac{998}{1000}+\frac{997}{1000}+...+\frac{1}{1000}$
