Tính tổng:
S = \(\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+...+\dfrac{2018}{2017.2018}\)
Những câu hỏi liên quan
tính tổng:\(\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+......+\dfrac{2018}{2017.2018}\)
\(\dfrac{2018}{6}+\dfrac{2018}{12}+...+\dfrac{2018}{2017.2018}\\ =2018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\right)\\ =2018\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)\\ =1009\)
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Đặt \(A=\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+..+\dfrac{2018}{2017.2018}\)
\(A=2018\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2017.2018}\right)\)
\(A=2018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2017.2018}\right)\)
\(A=2018\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(A=2018\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)\)
\(A=2018\left(\dfrac{1006}{2018}-\dfrac{1}{2018}\right)=2018.\dfrac{1005}{2018}=1005\)
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so sánh
1) A = \(\dfrac{10^{11}-1}{10^{12}-1}\) và B =\(\dfrac{10^{10}+1}{10^{11}+1}\)
2) A = \(\dfrac{2018^9+1}{2018^{10}-1}\) và B = \(\dfrac{2018^{19}+1}{2018^{20}+1}\)
3) A = \(\dfrac{2018^{19}+1}{2018^{20}+1}\) và B = \(\dfrac{2018^{20}+1}{2018^{21}+1}\)
Với x\(\ne-1\) \(\left(\dfrac{x^2+2x+2}{x+1}\right)^{2018}=a_0+a_1x+a_2x^2+...+a_kx^{2018}+\dfrac{b_1}{x+1}+\dfrac{b_2}{\left(x+1\right)^2}+...+\dfrac{b_{2018}}{\left(x+1\right)^{2018}}.\). Tính: S=\(\sum\limits^{2018}_{k=1}bx\)
Đề bài: So sánh
1, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}với\) 3
2, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}với\dfrac{2017+2018}{2018+2019}\)
Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
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Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
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Chứng minh: \(A=1.2.3.....2017.2018\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)⋮2019\)
\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)
\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)
\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).
Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.
Vậy A chia hết cho 2019.
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so sánh P và Q , bt P =\(\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\) và Q= \(\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Giải:
Ta có:
\(P=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
và \(Q=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Vì \(\left\{{}\begin{matrix}\dfrac{2016}{2017}=\dfrac{2016}{2017}\\\dfrac{2017}{2018}=\dfrac{2017}{2018}\\\dfrac{2018}{2019}=\dfrac{2018}{2019}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Hay \(P=Q\)
Vậy ...
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BT1: So sánh:
2) \(\dfrac{2017}{2018}+\dfrac{2018}{2019}\) VỚI \(\dfrac{2017+2018}{2018+2019}\)
Ta có :
\(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Ta thấy :
\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\left(1\right)\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\left(2\right)\)
từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
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So sánh A và B , biết
\(A=\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}\)
\(B=\dfrac{2017+2018+2019}{2018+2019+2020}\)
Ta có: \(B=\dfrac{2017+2018+2019}{2018+2019+2020}=\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2019+2020}\)
Mà \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019+2020}\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019+2020}\)
\(\dfrac{2019}{2020}>\dfrac{2019}{2018+2019+2020}\)
\(\Rightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}>\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2919+2020}\)
\(\Rightarrow A>B.\)
Vậy \(A>B.\)
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Cho A = \(\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
B = \(\dfrac{2017+2018}{2018+2019}\)
So sánh A và B
Ta có: \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\)
=> \(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
=> A > B
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Ta có :
\(B=\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Ta thấy :
\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\left(1\right)\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A>B\)
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Ta có:
\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2016+2019}\) (2)
Từ (1) và (2)⇒ A> B
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