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\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{-1}{2}\right).\left(\dfrac{-2}{3}\right).\left(\dfrac{-3}{4}\right)...\left(\dfrac{-99}{100}\right)\) ( 99 phân số )
\(=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)...\left(-99\right)}{2.3.4...100}=\dfrac{-1}{100}\)
Vậy \(A=\dfrac{-1}{100}\)
\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)
Nhận xét
thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10
\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)
\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
Giải:
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)
\(=\dfrac{1.2.3...99}{2.3.4...100}\)
\(=\dfrac{1}{100}\)
Vậy ...
\(B=1+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+\dfrac{1}{4}.\left(1+2+3+4\right)+...+\dfrac{1}{100}.\left(1+2+3+...+100\right)\)
\(B=1+\dfrac{1}{2}.2.3:2+\dfrac{1}{3}.3.4:2+\dfrac{1}{4}.4.5:2+...+\dfrac{1}{100}.100.101:2\)
\(B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{101}{2}\)
\(B=\dfrac{2+3+4+...+101}{2}\)
Tự tính :v
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).........................\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\left(\dfrac{1}{4}-\dfrac{4}{4}\right)................\left(\dfrac{1}{99}-\dfrac{99}{99}\right)\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)
\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)
\(A=\dfrac{-1}{100}\)
Ta có
A = \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)(99 thừa số)
A = \(\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
A = \(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right).\left(-100\right)}{2.3.4....98.99.100}\)
A = \(\dfrac{\left(-1\right).\left(-1\right).\left(-1\right)....\left(-1\right)}{1.1.1...1.1.1}\) (100 số -1, 99 số 1)
A = \(\dfrac{-1}{1.1.1.1...1.1.1}\)
A = \(\dfrac{-1}{1}\)
A = -1
Vậy A = -1