Tính dùm mk cái nhé
\(\dfrac{\dfrac{1}{1003}+\dfrac{1}{1004}+...+\dfrac{1}{2017}}{\dfrac{1}{1003.2017}+\dfrac{1}{1004.2016}+...+\dfrac{1}{2017.1003}}\)
Những câu hỏi liên quan
Tìm x biết: \(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)
<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)
<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)
vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017
vậy..........
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Giải phương trình:
\(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\left(1\right)\)
\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)
\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)
\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)
Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
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chứng minh rằng
\(\dfrac{1}{1000}+\dfrac{1}{1002}+\dfrac{1}{1004}+...+\dfrac{1}{2000}< \dfrac{1}{2}\)
ủa bạn ơi, lớn hơn 1/2 hay bé hơn 1/2 vậy bạn
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24 tháng 2 2022 lúc 11:03
bài 1 : tính
\(B=\dfrac{1}{10.9}+\dfrac{1}{18.13}+\dfrac{1}{26.27}+...+\dfrac{1}{802.405}\)
\(D=\dfrac{1}{2}-\dfrac{1}{2^4}+\dfrac{1}{2^7}-\dfrac{1}{2^{10}}+...-\dfrac{1}{2^{58}}\)
mn giúp mk gấp nhé !!
20 tháng 4 2022 lúc 22:36
giúp mk, please :)
\(\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{2017+\dfrac{2016}{6}+\dfrac{2015}{7}+...+\dfrac{1}{2021}}\)
A. \(\dfrac{1}{2020}\)
B. \(\dfrac{1}{2021}\)
C. \(\dfrac{1}{2019}\)
D. \(\dfrac{1}{2022}\)
chọn ra 3 ngừi nhanh nhứt:>>
giải thích cho những ng ko hỉu ;-;
\(=\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{\left(\dfrac{2016}{6}+1\right)+\left(\dfrac{2015}{7}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{\dfrac{2022}{6}+\dfrac{2022}{7}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}}\)
\(=\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{2022.\left(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}\right)}=\dfrac{1}{2022}\)
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Xem thêm câu trả lời
Bài 1: Cho P 7+72+73+74+.........+72016. Chứng minh P chia hết cho 400.
Bài 2: Tìm giá trị lớn nhất
a) A | x - 1004 | - | x+1003 |
b) B | x - 2018 | - | x - 2017 |
Bài 3 : Cho dfrac{2x-4y}{3}dfrac{4z-3y}{2}dfrac{3y-2z}{4} . Tìm x,y,z biết 2x-y+z 27
Bài 4: Tìm các số thực x,y,z biết dfrac{x+y-3}{z}dfrac{y+z+1}{x}dfrac{x+z+2}{y}dfrac{1}{x+y+z}
Bài 5 : a) Tính : dfrac{1}{1.3}+dfrac{1}{3.5}+dfrac{1}{5.7}+.....+dfrac{1}{19.21}
b) Chứng minh : dfrac{1}{1.3}+dfrac{1}{3.5}+...+dfrac{1}{left(2n-1r...
Đọc tiếp
Bài 1: Cho P= 7+72+73+74+.........+72016. Chứng minh P chia hết cho 400.
Bài 2: Tìm giá trị lớn nhất
a) A= | x - 1004 | - | x+1003 |
b) B = | x - 2018 | - | x - 2017 |
Bài 3 : Cho \(\dfrac{2x-4y}{3}=\dfrac{4z-3y}{2}=\dfrac{3y-2z}{4}\) . Tìm x,y,z biết 2x-y+z = 27
Bài 4: Tìm các số thực x,y,z biết \(\dfrac{x+y-3}{z}=\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{1}{x+y+z}\)
Bài 5 : a) Tính : \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{19.21}\)
b) Chứng minh : \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n-1\right)}\) < \(\dfrac{1}{2}\)
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
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sắp xếp :
\(\dfrac{1001}{2002};\dfrac{-1003}{2003};\dfrac{-1002}{2003};\dfrac{-1003}{-2002};\dfrac{1004}{-2003}\)
ta thấy : \(\dfrac{-1003}{-2002}\) = \(\dfrac{1003}{2002}\)
\(\dfrac{1004}{-2003}\) = \(\dfrac{-1004}{2003}\)
Sắp xếp : \(\dfrac{1004}{-2003}\) <\(\dfrac{-1003}{2003}\) <\(\dfrac{-1002}{2003}\) <\(\dfrac{1001}{2002}\) <\(\dfrac{-1003}{-2002}\)
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A=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}\)
B=\(\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}\)
Tính \(^{\left(A-B\right)^{2019}}\)
25 tháng 9 2017 lúc 20:53
Cho \(\left\{{}\begin{matrix}m,n>0\\x^2+y^2=1\\\dfrac{x^2}{m}+\dfrac{y^2}{n}=\dfrac{1}{m+n}\end{matrix}\right.\)
CMR \(\dfrac{x^{1005}}{m^{1004}}+\dfrac{y^{1005}}{n^{1004}}=\dfrac{1}{\left(m+n\right)^{1004}}\)