\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\cdot\cdot\cdot\frac{10^2}{10\cdot11}\)
Những câu hỏi liên quan
\(A=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{9^2}{9\cdot10}\)
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}=\frac{1^2.2^2.3^2...9^2}{1.2.2.3.3.4.4...9.10}=\frac{1.2^2.3^2...9^2}{1.2^2.3^2.4^2...10^2}=\frac{1}{10^2}=\frac{1}{100}\)
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\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
tính GTBT:
N=\(\frac{-1^2}{1\cdot2}\cdot\frac{-2^2}{2\cdot3}\cdot\frac{-3^2}{3\cdot4}\cdot\cdot\cdot\frac{-100^2}{100\cdot101}\cdot\frac{-101^2}{101\cdot102}\)
\(N=\frac{-1^2}{1.2}.\frac{-2^2}{2.3}.\frac{-3^2}{3.4}....\frac{-100^2}{100.101}.\frac{-101^2}{101.102}\)
\(=\frac{1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}....\frac{100.100}{100.101}.\frac{101.101}{101.102}\)
\(=\frac{1.2.2.3.3....100.100.101.101}{1.2.2.3.3.4....100.101.101.102}\)
\(=\frac{1}{102}\)
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\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
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\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
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\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{999^2}{999\cdot1000}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{999^2}{999.1000}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.....\frac{999.999}{999.1000}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)
\(=\frac{1}{1000}\)
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G=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{50^2}{49.51}\)
H=\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{10}{7}\right)\)
Giúp mình vs
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
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\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{51}=\frac{100}{51}\)
\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)
\(=0\)
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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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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
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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