Đặt A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{101.102}\)
B=\(\frac{1}{52.102}+\frac{1}{53.101}+...+\frac{1}{102.52}\)
CMR: A/B thuộc Z
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30 tháng 9 2023 lúc 15:00
tính tỉ số \(\dfrac{A}{B}\) biết A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{101.102}\) và B=\(\dfrac{1}{52.102}\)+\(\dfrac{1}{53.101}\)+...+\(\dfrac{1}{102.52}\)+\(\dfrac{2}{77.154}\)
tìm x biết (1/1.2+1/3.4+1/5.6+...+1/101.102).x=1/52.102+1/53.101+...+1/102.52+1/77.154
cho A=1/1.2+1/3.4+1/5.6+......+1/101.102
B=1/52.102+1/53.102+1/54.100+......+1/101.53+1/102.52
cmr A/B là số nguyên
Mình mới lớp 4 mà bạn tra mạng sẽ có đầy bài dạng này và y như thế này.
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cho A=1/1.2+1/3.4+1/5.6+.....+1/101+102
B=1/52.102+1/53.101+1/54.100+.....+1/101.53+102.52
Chứng minh rằng :A/B là số nguyên
+ \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{102}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{102}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{102}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{102}\right)-\left(1+\frac{1}{2}+...+\frac{1}{51}\right)\)
\(A=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)
+ \(154B=\frac{52+102}{52\cdot102}+\frac{53+101}{53\cdot101}+...+\frac{102+52}{102\cdot52}\)
\(154B=\frac{1}{52}+\frac{1}{102}+\frac{1}{53}+\frac{1}{101}+...+\frac{1}{101}+\frac{1}{53}+\frac{1}{102}+\frac{1}{52}\)
\(154B=2\left(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\right)\)
\(B=\frac{1}{77}\left(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\right)\)
Do đó : \(\frac{A}{B}=\frac{1}{\frac{1}{77}}=77\) là số nguyên
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Tính tổng hoặc hiệu sau:Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+frac{1}{4.5}+..................+frac{1}{100.101}+frac{1}{101.102}Bfrac{1}{1.2}-frac{1}{2.3}-frac{1}{3.4}-frac{1}{4.5}- .....................-frac{1}{100.101}-frac{1}{101.102}
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Tính tổng hoặc hiệu sau:
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+..................+\(\frac{1}{100.101}\)+\(\frac{1}{101.102}\)
B=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\)-\(\frac{1}{4.5}\)- .....................-\(\frac{1}{100.101}\)-\(\frac{1}{101.102}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
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A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
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B=1/1.2-1/2.3-1/3.4-1/4.5-.......1/100.101-1/101.102
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/100-1/101+1/101-1/102
B=1-1/102
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CMR:
a) \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
b) Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
CMR: \(\frac{7}{12}< A< \frac{5}{6}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)
\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)
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a) Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013+2014}\); Đặt B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(B=\frac{1}{1010.2018}+\frac{1}{1011.2017}+...+\frac{1}{2018.1010}\)
CMR: \(\frac{A}{B}\in Z\)
Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2017.2018}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\)
\(3028B=\frac{1010+2018}{1010.2018}+\frac{1011+2017}{1011.2017}+..+\frac{2018+1010}{2018.1010}\)
\(=(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+...+\frac{1}{1010})+(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018})\)
\(=2A\)
\(\Rightarrow \frac{A}{B}=1514\in \mathbb{Z}\)
Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2017.2018}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\)
\(3028B=\frac{1010+2018}{1010.2018}+\frac{1011+2017}{1011.2017}+..+\frac{2018+1010}{2018.1010}\)
\(=(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+...+\frac{1}{1010})+(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018})\)
\(=2A\)
\(\Rightarrow \frac{A}{B}=1514\in \mathbb{Z}\)
a) Chứng tỏ rằng
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\); B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên