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\(D=\frac{1}{5}+\frac{1}{5^3}+....+\frac{1}{5^{101}}-\frac{1}{5^2}-\frac{1}{5^4}-....-\frac{1}{5^{100}}\)
\(5D=5+\frac{1}{5^2}+...+\frac{1}{5^{100}}-\frac{1}{5}-....-\frac{1}{5^{99}}\)
\(5D+D=5+\frac{1}{5^2}+...+\frac{1}{5^{100}}-\frac{1}{5}-....-\frac{1}{5^{99}}+\frac{1}{5}+...+\frac{1}{5^{101}}-\frac{1}{5^2}-....-\frac{1}{5^{100}}=5+\frac{1}{5^{101}}\)
\(D=\frac{5^{102}+1}{5^{101}}\div6\)
ok men
Ta có : \(A=\frac{1}{1.101}+\frac{1}{2.202}+\frac{1}{3.103}+...+\frac{1}{10.110}\)
=\(\frac{1}{100}.\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right)\)
= \(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right)\)
= \(\frac{1}{100}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)
Lại có : B = \(\frac{1}{10}.\left(\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\right)\)
= \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\right)\)
= \(\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)
Khi đó \(A:B=\frac{A}{B}=\frac{\frac{1}{100}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}{\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}=\frac{1}{10}\)
a) = 3/3 x ( -24/54 +45/54 ) : 7/12
= 1 x 21/54 x 12/7
= 18/27
( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )
S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)
S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)
S = \(\frac12\) - \(\frac19\)
S = \(\frac{9}{18}-\frac{2}{18}\)
S = \(\frac{7}{18}\)
a; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2
1 = 1 = 1
1/2^2 < 1/1.2 = 1/1 - 1/2
1/3^2 < 1/2.3 = 1/2 - 1/3
.......................
1/100^2 < 1/99.100 = 1/99 - 1/100
Cộng vế với vế ta có:
A = 1 + 1 - 1/100
A = 2 - 1/100 < 2 (đpcm)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
I Don't Know OK