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S = 2(5/1.6 + 5/6.11 +.......+ 5/101.106)
S = 2( 1 - 1/6 + 1/6 - 1/11 +.....+ 1/101 - 1/106)
S = 2( 1 - 1/106)
S = 2 . 105/106
S = 105/53
k mk đi,mk mới bị trừ điểm!
1/2.S=5/(1.6)+5/(6.11)+...+5/(101.106)
1/2.S=1/1-1/6+1/6-1/11+...+1/101-1/106
1/2.S=1/1-1/106
1/2.S=105/106
S=105/53
\(S=2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)
\(S=2\left(1-\frac{1}{106}\right)\)
\(S=\frac{210}{106}=\frac{105}{53}\)
\(S=2.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)
\(S=2.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{106}\right)\)
\(S=2.\left[1+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{11}\frac{1}{11}\right)+...+\left(\frac{-1}{101}+\frac{1}{101}\right)-\frac{1}{106}\right]\)
\(S=2.\left[1+0+0+...+0-\frac{1}{106}\right]\)
\(S=2.\left[1-\frac{1}{106}\right]\)
\(S=2.\frac{105}{106}\)
A=1/15-1/16+1/16-1/17+...+1/2016-1/2017
A=1/15-1/2017
A=2002/30255
C=1/3[3/5.8+3/8.11+...+3/101.104]
C=1/3[1/5-1/8+1/8-1/11+...+1/101-1/104]
C=1/3[1/5-1/104]
C=1/3.99/520
C=33/520
\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{96.101}\\ S=\dfrac{25}{1.6}+\dfrac{25}{6.11}+\dfrac{25}{11.16}+...+\dfrac{25}{96.101}\\ S=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right)\\ S=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ S=5.\left(1-\dfrac{1}{101}\right)\\ S=5.\dfrac{100}{101}\\ S=\dfrac{500}{101}\)
P= \(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.17}\)+...+\(\dfrac{3}{96.101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{5}{3}\).(\(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+...+\(\dfrac{3}{96.101}\))
\(\dfrac{5}{3}\).P= \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+...+\(\dfrac{5}{96.101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{16}\)+...+\(\dfrac{1}{96}\)-\(\dfrac{1}{101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{101}\)= \(\dfrac{101}{101}\)-\(\dfrac{1}{101}\)=\(\dfrac{100}{101}\)
P= \(\dfrac{100}{101}\):\(\dfrac{5}{3}\)= \(\dfrac{100}{101}\).\(\dfrac{3}{5}\)=\(\dfrac{100.3}{101.5}\)=\(\dfrac{20.3}{101.1}\)=\(\dfrac{60}{101}\)
Vậy P= \(\dfrac{60}{101}\)
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
\(a,\frac{20132013}{20142014}=\frac{2013.10001}{2014.10001}=\frac{2013}{2014}=1-\frac{1}{2014};\frac{131313}{141414}=\frac{13.10101}{14.10101}=\frac{13}{14}=1-\frac{1}{14}.\text{Vì: 14 bé hơn 2014 nên:}\frac{1}{14}>\frac{1}{2014}\Rightarrow\frac{20132013}{20142014}>\frac{131313}{141414}\)
\(C=2013^9+2013^9.2013=2013^9\left(2013+1\right)=2013^9.2014;D=2014^9.2014\text{ vì: 2013^9< 2014^9 nên: C bé thua D }\)
\(c,M=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}};N=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2005}}.Vì:10^{2006}>10^{2005}.Nên:\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\Rightarrow M>N\)
Tính S:
S=5.(\(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+...+\(\dfrac{5}{101.106}\))
S=5.(1-\(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\))
S=5.(1-\(\dfrac{1}{106}\))
S=5.\(\dfrac{105}{106}\)
S=\(\dfrac{525}{106}\)
525/106