Tính A = \(\frac{1}{1\times2}\)+ \(\frac{1}{2\times3}\)+\(\frac{1}{3\times4}\)+ ... +\(\frac{1}{18\times19}\)+\(\frac{1}{19\times20}\)?
Những câu hỏi liên quan
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{18\times19}+\frac{2}{19\times20}\)
Tính nhanh
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+.......+\frac{2}{18.19}+\frac{2}{19.20}.\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=\frac{2.19}{20}=\frac{19}{10}\)
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\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...-\frac{1}{19}+\frac{1}{20}\)
\(=1+\frac{1}{20}\)
\(=\frac{1}{20}\)
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\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+......+\frac{1}{19\times20}\)
\(=2\left(\frac{1}{1\times2}+\frac{1}{2\times3}+..........+\frac{1}{19\times20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
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Xem thêm câu trả lời
Chứng minh rằng: \(\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+....+\frac{1}{19\times19}+\frac{1}{20\times20}
\(\frac{1}{2\times2}+\frac{1}{3\times3}+....+\frac{1}{20\times20}
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Cho biểu thức A= \(\frac{1}{1\times2\times3}\)+ \(\frac{1}{2\times3\times4}\)+ \(\frac{1}{3\times4\times5}\)+...+ \(\frac{1}{18\times19\times20}\). So sánh A với \(\frac{1}{4}\).
Cho biểu thức A= 11×2×3 + 12×3×4 + 13×
4×5 +...+ 118×19×20 . So sánh A với 14 .
Dương Đình Hưởng
cố lên mà k
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Chứng tỏ rằng:\(\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+...+\frac{39}{19^2\times20^2}\)< 1
\(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{19^2x20^2}\)<1
=\(\frac{3}{1.4}\)+\(\frac{5}{4x9}\)+...+\(\frac{39}{361x400}\)<1
=1-\(\frac{1}{4}\)+\(\frac{1}{4}\)-...-\(\frac{1}{361}\)+\(\frac{1}{361}\)-\(\frac{1}{400}\)<1
vì 1-\(\frac{1}{400}\)<1 nên \(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{39^2x40^2}\)<1
vậy..............................................
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Thực hiện phép tính:a)1times2times3times......times9-1times2times3times......times8-1times2times3times.....times8times8b)Bfrac{left(3times4times2^{16}right)^2}{11times2^{13}times4^{11}-16^9}c)C70timesleft(frac{131313}{565656}+frac{131313}{727272}+frac{131313}{909090}right)d)Bfrac{1}{4times9}+frac{1}{9times14}+frac{1}{14times19}+...+frac{1}{64times69}GIÚP MIK VỚI AI NHANH MIK CHO 3 TICK LUÔN CHO NHA!!!!!!!!!!NHỚ TRÌNH BÀY RÕ RÀNG NHA!!!!!!HELP ME
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Thực hiện phép tính:
a)\(1\times2\times3\times......\times9-1\times2\times3\times......\times8-1\times2\times3\times.....\times8\times8\)
b)\(B=\frac{\left(3\times4\times2^{16}\right)^2}{11\times2^{13}\times4^{11}-16^9}\)
c)\(C=70\times\left(\frac{131313}{565656}+\frac{131313}{727272}+\frac{131313}{909090}\right)\)
d)\(B=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{64\times69}\)
GIÚP MIK VỚI AI NHANH MIK CHO 3 TICK LUÔN CHO NHA!!!!!!!!!!NHỚ TRÌNH BÀY RÕ RÀNG NHA!!!!!!HELP ME
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
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d)\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\right)\)
\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{69}\right)\)
\(\frac{1}{3}\times\frac{65}{276}\)
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Bài 1 : Tính tổng S , biết : \(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{2010\times2011}\)
Bài 2 : Tính tổng sau : \(S=\frac{3}{10\times13}+\frac{3}{13\times16}+\frac{3}{16\times19}+....+\frac{3}{58\times61}\)
Bài 3 : Tính tổng sau : \(S=\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}+....+\frac{1}{19\times22}\)
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
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Tính\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...\frac{1}{2014\times2015\times2016}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)
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Cho B= \(\frac{1\times2}{1\times2\times3}+\frac{1\times2}{1\times2\times4}+\frac{1\times2}{1\times2\times3\times4}+\frac{1\times2}{1\times2\times3\times4\times5}+....+\frac{1\times2}{n,giao}\left(n\in N,n\ge3\right)\)
chứng tỏ B nhỏ hơn 3
tính :\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\frac{1}{4\times5\times6}+\frac{1}{5\times6\times7}+\frac{1}{6\times7\times8}+\frac{1}{7\times8\times9}+\frac{1}{8\times9\times10}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
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