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\(P=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}\right)-\left(\frac{1}{98}-\frac{1}{97}\right)-\left(\frac{1}{97}-\frac{1}{96}\right)-...-\left(\frac{1}{3}-\frac{1}{2}\right)-\frac{1}{2}\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-\frac{1}{97}+\frac{1}{96}-...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}\)
\(=0\)
ĐS: \(0\)
=\(\frac{1}{99}\)-\(\frac{1}{99}\)-\(\frac{1}{98}\)-\(\frac{1}{98}\)-.................-\(\frac{1}{3}\)-\(\frac{1}{2}\)-\(\frac{1}{2}\)-1
=\(\frac{1}{99}\)-(\(\frac{1}{99}\)+\(\frac{1}{98}\)+..............+\(\frac{1}{3}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+1)
=\(\frac{1}{99}\)-......
hình như sai rùi????
D =1/99 -1/99.98-1/98.97-...-1/3.2-1/2.1
=1/99-(1/99.98+1/98.97-...-1/3.2+1/2.1)
=1/99-(1/1.2+1/2.3+1/3.4+...+1/98.99)
=1/99-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-...+1/98-1/99)
=1/99-(1/1-1/99)
=1/99-98/99
=-97/99
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(A=\frac{1.1.2.2.3.3...9.9}{1.2.2.3.3.4...9.10}\)
\(A=\frac{1}{10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-1\right)\)
\(B=\frac{1}{99}-\frac{1}{99}+1\)
\(B=1\)
S=\(\frac{1}{100}-\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-......-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)\(=1-\frac{1}{100}-\frac{2}{99}\)\(=\frac{9601}{9900}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}=...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right).\)
\(=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
chúc bạn học tốt
b) \(GọiB=\frac{-1}{100.99}+\frac{-1}{99.98}+...+\frac{-1}{2.1}\)
\(2B=\frac{-2}{100.99}+\frac{-2}{99.98}+...+\frac{-2}{2.1}\)
\(2B=\frac{-1}{100}-\frac{-1}{99}+\frac{-1}{99}-\frac{-1}{98}+...+\frac{-1}{2}-\frac{-1}{1}\)
\(2B=\frac{-1}{100}-\frac{-1}{1}\)
\(2B=\frac{99}{100}\Rightarrow B=\frac{99}{100}:2=\frac{99}{200}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(< =>3A=\frac{3}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{3}{3^8}\)
\(< =>3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(< =>3A-A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
\(< =>A=\frac{3^8-1}{\frac{3^8}{2}}\)
\(D=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(D=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(D=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(D=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{99}\right)\)
\(D=\frac{1}{99}-\frac{1}{100}-1+\frac{1}{99}\)
b tự làm nốt nhé
D=1100.99 −199.98 −198.97 −...−13.2 −12.1
D=199.100 −(11.2 +12.3 +...+197.98 +198.99 )
D=199 −1100 −(1−12 +12 −13 +...+198 −199 )
D=199 −1100 −(1−199 )
D=199 −1100 −1+199
=1\99-(1\1.2+1\1.3+....1\98.99)
=1\99-(1-1\2+1\2-1\3+1\3...+1\98-1\99)
=1\99-(1-1\99)
=1\99-1\98
=-97\99
\(P=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{98.99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
Vậy \(P=-\frac{97}{99}\)
P=-1/1.2-1/2.3-...-1/98.99-1/99
P=-(1/1.2+1/2.3+...+1/98.99+1/99)
P=-1