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\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{49.50}\)
=1-\(\frac{1}{2}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
=(1+\(\frac{1}{3}\)+\(\frac{1}{5}\)+...+\(\frac{1}{49}\))-(\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{6}\)+...+\(\frac{1}{50}\))
=(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+...+\(\frac{1}{50}\))-2(\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{6}\)+...+\(\frac{1}{50}\))
=(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+...+\(\frac{1}{50}\))-(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+...+\(\frac{1}{25}\))
=\(\frac{1}{26}\)+\(\frac{1}{27}\)+\(\frac{1}{28}\)+...+\(\frac{1}{50}\)\(\Rightarrow\)ĐPCM
1/1.2 + 1/3.4 + 1/5.6 +.....+1/49.50
=1- 1/2 + 1/3 - 1/4 +1/5 -1/6+....+1/49 -1/50
=(1 +1/3 +1/5 +....+1/49) - (1/2 +1/4 +1/6 +....+1/50)
=(1+1/2 +1/3 +....+1/50) - 2(1/2 + 1/4 + 1/6 +....+ 1/50)
=1+1/2 +1/3 +.....+1/50 - (1 +1/2 +1/3 +.....+1/25)
=1+1/2 +1/3 +....+1/50 -1-1/2-1/3-...-1/25
=1/26+ 1/27 +1/28 +....+1/50
Vậy 1/1.2 + 1/3.4 + 1/5.6 + .....+ 1/49.50=1/26 + 1/27 + 1/28 + ....+1/50
Mình thấy bài này dễ mà, quên mất , mình là học sinh lớp 6 đấy. Bài này như kiểu toán nâng cao lớp 6 ý. Mình nghĩ đây ko phri toán lớp 7 đâu.
A=1/1.2+1/3.4+1/5.6+....+1/99.100
A=2-1/1.2+4-3/3.4+6-5/5.6+...+100-99/99.100
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+....+1/99-1/100
A=1-1/100
A=99/100
Cmr:7/12<A<5/6
7/12<99/100<5/6
A=1/1.2+1/3.4+...+1/99.100=1/51+1/52+...+1/100
=>A>1/75.25+1/100.25=1/3+1/4=7/12
=>A>7/12
A<1/51.25+1/76.25<1/50.25+1/75.25=1/2+1/3=5/6
=>A<5/6
Vậy:7/12<A<5/6
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)
=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)<\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{75}+...+\frac{1}{75}\right)\)
=> \(A<\frac{25}{50}+\frac{25}{100}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}<\frac{5}{6}\)
Vậy...
Chứng minh : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\): như câu trên
a) S1 = \(-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{99.100}\)
= \(-\frac{1}{1}-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{99}-\frac{1}{100}\)
= \(\frac{-1}{1}-\frac{1}{100}\)
= \(-\frac{101}{100}\)
a) Để A thuộc Z thì 3 phải chia hết cho n-1
=> n-1 thuộc Ư(3)={1;3;-1;-3}
=> n thuộc {2;4;0;-2}
b) ta có : A=(6n+5)/(2n-1)=[3(2n-1)+8]/(2n-1)=3+[8/(2n-1)]
Để A thuộc Z thì 8 chia hết cho 2n-1
=>2n-1 thuộc Ư(8)={1;2;4;8;-1;-2;-4;-8}
=>2n thuộc { 2;0}
=> n thuộc {1;0}
Câu c và bài 2 bạn tự làm đi nghe
Bạn nên đổi chử thuộc và chia hết thành đấu nghe
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{49}+\frac{1}{50}\right)\)
\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Nhận xét :
\(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)
=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)(Đề bài của bạn đánh sai)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)<\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{75}+...+\frac{1}{75}\right)\)
=> \(A<\frac{25}{50}+\frac{25}{75}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
=> ĐPCM
Ta thấy : A = 1 - 1/2 + 1/3 - 1/4 +...+ 1/99 +1/100
A = 1 + 1/2 + 1/3 + 1/4 +...+ 1/99 + 1/100 - 2. (1/2 + 1/4 +1/6 +...+ 1/100)
A = 1 + 1/2 + 1/3 +1/4 +...+ 1/99 + 1/100 - (1 + 1/2 + 1/3 +...+1/50)
A = 1/51 + 1/52 + 1/53 +...+ 1/100
Do đó : A = (1/51 + 1/52 + 1/53 +...+ 1/75) + (1/76 + 1/77 +...+ 1/100)
Ta có : 1/51 > 1/52 > ... > 1/75 ; 1/76 > 1/77 > ... > 1/100 nên :
A > 1/75.25 + 1/100.25 = 1/3 + 1/4 = 7/12
A < 1/51.25 < 1/50.25 + 1/75.25 = 1/2 +1/3 = 5/6
Vậy 7/12 < A <5/6 . ^_^
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(=1-\frac{1}{2000}\)
\(=\frac{1999}{2000}\)
1/1.2+1/2.3+1/3.4+..+1/1999.2000
=1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/2000
=1-1/2000