so sánh \(\frac{1}{101}\)+ \(\frac{1}{102}\)+ ......+\(\frac{1}{199}\)+\(\frac{1}{200}\)với \(\frac{5}{8}\)
ai có cách giải mình cho 2 tick
K
Khách
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19 tháng 5 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
AK
19 tháng 5 2018
Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(đpcm\right)\)
Chúc bạn học tốt !!!
7 tháng 4 2017
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)+\left(\frac{1}{101}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{199}+\frac{1}{200}\) (ĐPCM)
7 tháng 4 2017
Ta có : 1 - 1/2 + 1/3 - 1/4 + ....- 1/200
= (1 + 1/3 + 1/5 + ....+ 1/199) - ( 1/2 + 1/4 + 1/6 + .... + 1/200)
= ( 1 + 1/3 +...+ 1/199) + (1/2 +1/4 + ...+ 1/200) - 2(1/2+1/4+...+ 1/200)
= (1+1/2+1/3+....+1/199 + 1/200) - (1 +1/2 +1/3 +....+1/100)
= 1/101 + 1/102+ 1/103 + .... + 1/200
chúc bạn học tốt!!!!!!!
S
27 tháng 5 2017
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
HN
27 tháng 5 2017
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
1 tháng 4 2016
Ta có : \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)\(\left(đpcm\right)\)
12 tháng 6 2020
Ta có : \(\frac{1}{101}>\frac{1}{200}\)
\(\frac{1}{102}>\frac{1}{200}\)
\(...>\frac{1}{200}\)
Mà \(\frac{1}{200}=\frac{1}{200}\)
Suy ra : \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)
Mời nhân tài giải nốt.
KN
13 tháng 4 2019
Tham khảo ở link này bạn nhé :
https://olm.vn/hoi-dap/detail/5631756599.html
~ Study well ~
27 tháng 2 2018
Biến đổi vế trái ta có :
\(VT=\frac{1}{1}+\frac{1}{3}+...+\frac{1}{199}+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)-\) \(2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}+...+\frac{1}{200}-\) \(1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\) \(=VP\RightarrowĐPCM\)
AM
27 tháng 6 2015
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Vậy \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{200}\)
ai nhanh nhất cho 1 k , nhanh nhé mình cần rất gấp chiều nay phải nộp rùi
có lẽ là bé hơn bn ạ