\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{10000}\right)\)
giúp em với các anh chị ơi
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
a.
\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)
\(=\left(-\frac{1}{2}\right)\times\left(-\frac{2}{3}\right)\times\left(-\frac{3}{4}\right)\times...\times\left(-\frac{2015}{2016}\right)\times\left(-\frac{2016}{2017}\right)\)
\(=\frac{1}{2017}\)
b.
\(\frac{2^{50}\times7^2+2^{50}\times7}{4^{26}\times112}=\frac{2^{50}\times\left(7^2+7\right)}{\left(2^2\right)^{26}\times112}=\frac{2^{50}\times\left(49+7\right)}{2^{52}\times2\times56}=\frac{56}{2^3\times56}=\frac{1}{8}\)
a. (1/2-1).(1/3-1)(1/4-1). ... .(1/2017-1)=(-1/2)(-2/3)(-3/4). ... .(-2016/2017)
Vì dãy số có 2016 số hạng âm nên tích của chúng là một số dương.
Ta có:(-1/2)(-2/3)(-3/4). ... . (-2016/2017)=1/2017
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
`Answer:`
\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{99.101}{100^2}\)
\(=\frac{\left(1.2.3.4.5...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4.5...100\right)}\)
\(=\frac{1}{100}.\frac{1}{2}.101\)
\(=\frac{101}{200}\)
\(=\frac{101}{200}\)
#zinc