Ví dụ 8: Tính một cách hợp lý: A= (1 - 1/1.2) + (1 - 1/2.3) +.....+(1- 1 2021.2022 )
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tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
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Tính:
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+,.......+\(\dfrac{1}{2021.2022}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
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\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
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1/1.2+1/2.3+1/3.4+…+1/2021.2022
-Nghỉ Tết đi :)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)=\(1-\dfrac{1}{2022}\)=\(\dfrac{2021}{2022}\)
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\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
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\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}\)
\(=\dfrac{2021}{2022}\)
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Tính tổng :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{2020.2021}+\dfrac{1}{2021.2022}\)
Dấu chấm là dấu nhân
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
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1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
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23 tháng 12 2022 lúc 21:19
tính hợp lý nếu có thể A=1.2+2.3+...+(2013.20.19)-1^2+2+3^2+...+2013^2)
Bài tập: Thực hiện phép tính sau bằng cách hợp lý nhất
[ 1/1.2 + 1/2.3 + 1/3.4 +........+ 1/199.200 ] : [1/101 + 1/102 + 1/103 +......+ 1/200 ]
Mong các bạn giúp mình giải bai này, mình cảm ơn trước nhé !
Tính hợp lý
a) A = \(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{2016.2017}\right)\)
Ta có : \(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+.......+\left(1-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=\left(1+1+1+......+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=2016-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(\Rightarrow A=2016-\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=2016-\frac{2016}{2017}=2015\frac{1}{2017}\)
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Tính bằng cách hợp lý
a) \(\frac{-1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{99.100}\)
b)\(\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{96.100}\)
Tính một cách hợp lí tổng sau :
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}.\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
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A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
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