Tính nhanh:
P = \(\frac{1}{99}\)-\(\frac{1}{99x98}\)-\(\frac{1}{98x97}\)-\(\frac{1}{97x96}\)-...-\(\frac{1}{3x2}\)-\(\frac{1}{2x1}\)
x = dấu nhân.
Gíup mình nhé.
K
Khách
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Những câu hỏi liên quan
LC
1
5 tháng 9 2015
Đặt A = \(\frac{1}{99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)
\(A=\frac{1}{99}-\left[-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{98.99}\right)\right]\)
\(A=\frac{1}{99}+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{99}\right)\)
\(A=\frac{1}{99}+\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{99}+\frac{98}{99}=1\)
NC
8 tháng 7 2020
Bài làm:
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(=\frac{1}{99.100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}\right)\)
\(=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99.100}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99.100}-\frac{98}{99}\)
\(=\frac{1-98.100}{99.100}=\frac{1-9800}{9900}=-\frac{9799}{9900}\)
Học tốt!!!!
F
8 tháng 7 2020
\(\left(\frac{1}{100.99}\right)-\left(\frac{1}{99.98}\right)-\left(\frac{1}{98.97}\right)-...-\left(\frac{1}{3.2}\right)-\left(\frac{1}{2.1}\right)\)
\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)
\(=\frac{1}{99}-\frac{1}{100}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1+\frac{1}{2}\right)\)
\(=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{1}{100}-1+\frac{1}{99}\)
\(=\frac{2}{99}-\frac{101}{100}\)
NT
3
N
25 tháng 8 2018
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=-\frac{98}{100}=-\frac{49}{50}\)
25 tháng 8 2018
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=\frac{1}{100}-\left(\frac{1}{100}-1\right)\)
\(=1\)
BS
2
22 tháng 5 2015
a,=(1/3+3/5+1/15)+(3/4+-1/36)+(1/72-2/9)=1+26/36-15/72=1+(52-15)/72=1+37/72=109/72
b,=1/100-(1/1x2+1/2x3+...+1/97x98+1/98x99+1/99x100)
=1/100-(1/1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
=1/100-(1/1-1/100)=1/100-99/100=-98/100=-49/50
chỉ có mk mk giải thôi đó l-i-k-e đi
K
30 tháng 5 2019
= \(\frac{11}{10}\cdot\frac{12}{11}\cdot\frac{13}{12}\cdot\frac{14}{13}\cdot\frac{15}{14}\cdot\frac{16}{15}\cdot\frac{17}{16}\)
=11/10 x 12/11 x 13/12 x 14/13 x 15/14 x 16/15 x 17/16
= \(\frac{17}{10}\)
BM
30 tháng 5 2019
=\(\frac{11}{10}\)x \(\frac{12}{11}\)x .......... x \(\frac{16}{15}\)x\(\frac{17}{16}\)
= \(\frac{11^1x12^1x......x16^1x17}{10x11^1x...x15^1x16^1}\)( những số có số nhỏ ở trên là rút gọn với số khác VD:11 rút gọn cho 11 )
=\(\frac{1x1x......x1x17}{10x1x.......x1x1}\)
=\(\frac{17}{10}\)
= 1,7
KN
6 tháng 5 2018
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
KN
7 tháng 5 2018
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
4 tháng 8 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)
\(=\frac{1999}{1000}\)
Tham khảo nhé~
29 tháng 6 2016
\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right)............\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}..............\frac{3}{2}\)
\(=\frac{101.100............3}{100.99...............2}\)
\(=\frac{101}{2}\)
VD
29 tháng 6 2016
\(=\frac{101}{100}.\frac{100}{989}.....\frac{3}{2}=\frac{101}{2}\)
T
29 tháng 7 2017
Đặt: \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(=\frac{1-\left[\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right]}{1-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}\)
\(=\frac{1-\frac{99}{1}}{1-\frac{1}{100}}\)
\(M=\frac{-98}{99}\)
Đặt \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
\(=\frac{92+\left[\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right]}{1-\left[\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right]}\)
\(=\frac{92+\frac{92}{100}}{1-\frac{1}{500}}\)
\(=\frac{92+\frac{92}{100}}{\frac{499}{500}}\)
Tự làm tiếp đi!
Ta có: \(P=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)
\(=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{2}{99}-1=\frac{2}{99}-\frac{99}{99}\)
\(=\frac{-97}{99}\)