Ta có:
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{4}{6}.\frac{9}{12}....\frac{9801}{9900}.\frac{10000}{10100}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}=\frac{1.2.3...99.100}{2.3.4...100.101}=\frac{1}{101}\)(Tối giản)
\(y=\frac{1\times100+2\times99+3\times98...+99\times2+100\times1}{1\times2+2\times3+3\times4+...+99\times100+100\times101}=?\)
\(A=\frac{1^2}{1\times2}\times\frac{2^2}{2\times3}\times\frac{3^2}{3\times4}\times\frac{4^2}{4\times5}\)
tính các tích sau
\(a=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{9999}{10000}\)
\(b=\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{10000}\right)\)
\(c=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(d=\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99\times100}\right)\)
Cho A=\(\frac{1\times2-1}{2!}+\frac{2\times3-1}{3!}+\frac{3\times4-1}{4!}+.....+\frac{99\times100-1}{100!}<2\)
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)\(B=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{64\times69}\)
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Chứng tỏ rằng: \(\frac{1}{1}\times\frac{1}{3}\times\frac{1}{5}\times.....\times\frac{1}{99}=\frac{2}{51}\times\frac{2}{52}\times\frac{2}{53}\times.....\times\frac{2}{100}\)
\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
tìm x biết
\(\frac{x\times\left(1\times2+2\times3+3\times4+...+98\times99\right)}{98\times100\times33}=2010-|-2011|\)
\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}...\frac{99}{100}\)
