Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
ta có 5^x+5^x*25=650 suy ra 5^x *26=650 ,5^x=25 suy ra x=2
Mk giúp phần đầu nhé!
Có 5x+5x+2=650
=> 5x.(1+52)=650
5x. 26=650
5x=650:26
5x=25
=> x=2
Ta có : \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}=\frac{1}{2}-\frac{2}{27}=\frac{23}{54}\)
Trả lời:
\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)
\(=\frac{1}{2}-\frac{2}{27}\)
\(=\frac{23}{54}\)
Học tốt
\(\frac{1}{100.99}-\frac{1}{99.98}-......-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(-\frac{1}{100.99}+\frac{1}{99.98}+...........+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(-\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+......+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(-\frac{1}{100}-1\right)\)
\(=\frac{1}{100}+1\)
\(=\frac{101}{100}\)
P=(1-1/1+2).(1-1/1+2+3)....(1-1/1+2+3+4+...+2011)
=[1-1/(2+1).2:2].[1-1/(3+1).3:2].....[1-1/(2011+1).2011:2]
=(1-2/2.3).(1-2/3.4)...(1-2/2011.2012)
=4/2.3.10/3.4....4046130/2011.2012
=1.4/2.3 .2.5/3.4 ....2010.2013/2011.2012
=1.2....2010/2.3...2011 .4.5....2013/3.4....2012
=1/2011.2013/3
=671/2011
Giúp mình với nhé.thanhk you!!!