\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{99}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
S
10 tháng 5 2017
Lời giải:\(A=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-4\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)
\(A=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
\(\Rightarrow A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right)}{2.3.4....99.100}\)
\(\Rightarrow A=\dfrac{1}{100}\)
20 tháng 7 2017
a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(=\dfrac{3.4.5...100}{2.3.4...99}\)
\(=\dfrac{100}{2}=50\)
MV
20 tháng 7 2017
a,
\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)
b,
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)
c,
\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)
17 tháng 7 2017
\(C=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{1-99^2}{100^2}\right)\left(\dfrac{100^2-1}{99^2}\right)=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{-2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{-98.100}{99^2}\right)\left(\dfrac{99.101}{100^2}\right)=-\dfrac{101}{200}\)
29 tháng 4 2021
Ta có: \(B=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 12 2023
A = (\(\dfrac{1}{2}\) + 1).(\(\dfrac{1}{3}\) + 1).(\(\dfrac{1}{4}\) + 1)...(\(\dfrac{1}{99}\) + 1)
A = \(\dfrac{1+2}{2}\).\(\dfrac{1+3}{3}\).\(\dfrac{1+4}{4}\)...\(\dfrac{1+99}{99}\)
A = \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\)....\(\dfrac{100}{99}\)
A = \(\dfrac{100}{2}\) \(\times\) \(\dfrac{3.4.5...99}{3.4.5...99}\)
A = 50
27 tháng 6 2022
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
8 tháng 5 2017
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Vậy T = 50
8 tháng 5 2017
\(T=\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{98}+1\right)\cdot\left(\dfrac{1}{99}+1\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}+\dfrac{4}{4}\right)\cdot...\cdot\left(\dfrac{1}{98}+\dfrac{98}{98}\right)\cdot\left(\dfrac{1}{99}+\dfrac{99}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{2\cdot3\cdot4\cdot...\cdot98\cdot99}\)
\(=\dfrac{100}{2}=50\).
Bằng \(\dfrac{1}{99}\)
\(...=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{97}{98}.\dfrac{98}{99}\)
\(=\dfrac{1}{99}\)