Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
A= (1/4-1)(1/9-1).......(1/10000-1)
A=-3/4(-8/9).........(-9999/100^2)
A=-1.3/2.2 (-2.4/3.3)........(-99.101/100.100)
A=-1.(-2).(-3)........(-99)/2.3.4......100 . 2.3.4......101/.3.4....100
A=-1/100 . 102/3=17/50
Vậy A= 17/50
THấy ảnh đại diện của mk thì tk và gửi kb với mk nha
Yêu các bạn nhìu(^-^)
\(B= \left(\frac{1}{2}-1\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}-1\right)\left(\frac{1}{4}+1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{100}+1\right)\)
\(B=-\frac{1}{2}.\frac{3}{2}.\frac{-2}{3}.\frac{4}{3}.\frac{-3}{4}.\frac{5}{4}.\frac{-4}{5}......\frac{-99}{100}.\frac{101}{100}\)
\(B=-\frac{1}{2}.\frac{101}{100}=-\frac{101}{200}\)
\(\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{10}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\frac{1}{100}\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot0\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)=0
\(\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).\left(\frac{1}{100}-\left(\frac{1}{2}\right)^2\right)......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)....\left(\frac{1}{100}-\left(\frac{1}{10}\right)^2\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)...\left(\frac{1}{100}-\frac{1}{100}\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).....0......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=0\)
\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(=-\frac{3}{4}.-\frac{8}{9}.\frac{-15}{16}...-\frac{9999}{10000}\)
\(=\left(-1\right)^{50}.\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(=1.\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(=\frac{3.8.15...9999}{4.9.16...10000}\)
\(=\frac{3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}\)
\(=\frac{\left(2.3.4...99\right).\left(3.4.5...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}\)
\(=\frac{1.101}{100.2}\)
\(=\frac{101}{200}\)
Tham khảo nha !!!
cảm ơn bạn rất nhiều