\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
Những câu hỏi liên quan
A = \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot.....\cdot\left(1+\frac{1}{2011\cdot2013}\right)\)
Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
Tìm n, biết:
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}>0,24995\)
tính \(A=\)\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot...\cdot\frac{99^2}{99\cdot100}\)
A = \(\frac{\left(1.2.3......99\right)\left(1.2.3......99\right)}{\left(1.2.3......99\right)\left(2.3.4.....100\right)}=\frac{1.1}{1.100}=\frac{1}{100}\)
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\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdot\cdot\cdot+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
A=1/1x2 +1/2x3 +... +1/18x19 + 1/19x20
Nhận xét 1/1x2 = 1/1 -1/2 ; 1/2x3=1/2-1/3; ... ;1/18x19=1/18-1/19 ; 1/19x20=1/19-1/20
ta có A=1/1 - 1/2 + +1/2 -1/3+1/3- +1/18-1/19+1/19-1/20
A=1/1 - 1/20
A=20/20 - 1/20
A=(20-1)/20
A=19/20
Vậy A=19/20
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A =\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ...+\(\frac{1}{18.19}\)+\(\frac{1}{19.20}\)
A = 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+....+\(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)
A = 1 - \(\frac{1}{20}\)( Vì đã triệt tiêu )
A = \(\frac{19}{20}\)
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a=(1/1-1/2) + (1/2 +1/3) +(1/3+1/4)+........+(1/2016+1/2017)
a=1/1-1/2+1/2-1/3+1/3-1/4+ ........+2016-2017
a=1/1+1/2017 =2017-1/2017
a=2016/2017
chac chan 100% do ban .tk mk nha
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Xem thêm câu trả lời
1) Tính
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
b)\(\left(\frac{15}{1\cdot2\cdot3}+\frac{15}{2\cdot3\cdot4}+\frac{15}{3\cdot4\cdot5}+.....+\frac{15}{18\cdot19\cdot20}\right)\cdot x=1\)
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
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a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33
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\(b,\)\(\left(\frac{15}{1.2.3}+\frac{15}{2.3.4}+\frac{15}{3.4.5}+...+\frac{15}{18.19.20}\right).x=1\)
\(\left[\frac{15}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{15}{18.19.20}\right)\right].x=1\)
\(\left[\frac{15}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right].x=1\)
\(\left[\frac{15}{2}.\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\right].x=1\)
\(\left[\frac{15}{2}.\frac{189}{380}\right].x=1\)
\(\frac{567}{152}.x=1\)
\(x=1-\frac{567}{152}\)
\(\Rightarrow x=-\frac{415}{152}\)
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tính:
A=\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}...\frac{8^2}{8\cdot9}\cdot\frac{9^2}{9\cdot10}\)
B=\(\frac{2^2}{3}\cdot\frac{^{3^2}}{8}\cdot\frac{4^2}{15}\cdot\frac{6^2}{35}\cdot\frac{7^2}{48}\cdot\frac{8^2}{63}\cdot\frac{9^2}{80}\)
A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)
A=\(\frac{1}{10}\)
mình làm đc 1 câu thôi. Bạn thông cảm nhé
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\(P=\frac{1}{1\cdot2\cdot3\cdot\text{4}}+\frac{1}{2\cdot3\cdot4\cdot5}+.........+\frac{1}{102\cdot103\cdot104\cdot105}\)