Rút gọn:
\(A=\frac{200+\frac{199}{2}+\frac{198}{3}+...+\frac{2}{199}+\frac{1}{200}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{200}+\frac{100}{201}}\)
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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: ...
SY
17 tháng 3 2016
Bạn ơi, của bạn giống của mình mà, đăng làm gì, nếu cùng chung 1 kết quả thì bạn tick đúng đi.
LH
2 tháng 1 2016
a, Dat A =\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\)
\(\Rightarrow\frac{1}{3}A+A=\left(\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\right)\)
\(\Rightarrow\frac{4}{3}A=\frac{1}{3}+\frac{1}{3^{200}}\)
\(\Rightarrow A=\frac{\frac{1}{3}+\frac{1}{3^{200}}}{\frac{4}{3}}\)
chung minh tuong tu cau b va c
S
17 tháng 3 2020
\(M=\left(1-\frac{1}{2}\right)-\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{6}\right)-....-\left(1-\frac{1}{200}\right)\)
\(M=-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-.....-\frac{1}{200}\right)=-\frac{1}{2}\left(1-\frac{1}{2}+...-\frac{1}{100}\right)\)
Xét:
\(S=1-\frac{1}{2}+....-\frac{1}{100}.S=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+....+\frac{1}{100}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\Rightarrow M=-\frac{1}{2}\left(\frac{1}{51}+....+\frac{1}{100}\right)\)
N:M=-2
U
25 tháng 7 2018
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\cdot\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right)}\)
\(=\frac{1}{200}\)
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