Thực hiện phép tính :
A = \(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot...\cdot\left(1-\dfrac{1}{1+2+3+...+2006}\right)\)
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
Rút gọn biểu thức
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)\)
Giúp em với ạ
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}\)
\(=\dfrac{10}{18}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{9}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{36234378}\)
Thực hiện tính
M=\(1+\dfrac{1}{2}\cdot\left(1+2\right)+\dfrac{1}{3}\cdot\left(1+2+3\right)+\dfrac{1}{4}\cdot\left(1+2+3+4\right)+...+\dfrac{1}{16}\cdot\left(1+2+3+...+16\right)\)
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
Thực hiện phép tính
\(A=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{1+2+3+..+2006}\right)\)
\(\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\cdot...\cdot\left(2005^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\cdot...\cdot\left(2006^4+\dfrac{1}{4}\right)}\).Viết kết quả dưới dạng phân số .Thanks!
Ta có: \(16a^4+4=16a^4+2.4a^2.2+4-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a^2-4a+2\right).\left(4a^2+4a+2\right)\)
\(=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( a \(\in\) N* )
Do đó: \(16a^4+4=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( * )
Thay a lần lượt bằng 1, 2, 3, ..., 2014, ta có:
\(16.1^4+4=\left[\left(2.1-1\right)^2+1\right].\left[\left(2.1+1\right)^2+1\right]=\left(1^2+1\right).\left(3^2+1\right)\)
\(16.2^4+4=\left[\left(2.2-1\right)^2+1\right].\left[\left(2.2+1\right)^2+1\right]=\left(3^2+1\right).\left(5^2+1\right)\)
\(16.3^4+4=\left[\left(2.3-1\right)^2+1\right].\left[\left(2.3+1\right)^2+1\right]=\left(5^2+1\right).\left(7^2+1\right)\)
\(16.4^4+4=\left[\left(2.4-1\right)^2+1\right].\left[\left(2.4+1\right)^2+1\right]=\left(7^2+1\right).\left(9^2+1\right)\)
\(......\)
\(16.2005^4+4=\left[\left(2.2005-1\right)^2+1\right].\left[\left(2.2005+1\right)^2+1\right]=\left(4009^2+1\right).\left(4011^2+1\right)\)
\(16.2006^4+4=\left[\left(2.2006-1\right)^2+1\right].\left[\left(2.2006+1\right)^2+1\right]=\left(4011^2+1\right).\left(4013^2+1\right)\)
Đặt \(T=\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)...\left(2005^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right)...\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{16.\left(1^4+\dfrac{1}{4}\right).16\left(3^4+\dfrac{1}{4}\right)...16\left(2005^4+\dfrac{1}{4}\right)}{16.\left(2^4+\dfrac{1}{4}\right).16\left(4^4+\dfrac{1}{4}\right)...16\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{\left(16.1^4+4\right).\left(16.3^4+4\right)...\left(16.2005^4+4\right)}{\left(16.2^4+4\right).\left(16.4^4+4\right)...\left(16.2006^4+4\right)}\)
\(\Leftrightarrow T=\dfrac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right)...\left(4009^2+1\right).\left(4011^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right)...\left(4011^2+1\right).\left(4013^2+1\right)}\)
\(\Leftrightarrow T=\dfrac{1^2+1}{4013^2+1}\)
\(\Leftrightarrow T=\dfrac{2}{4013^2+1}\)
Tính tổng:
\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{2019\cdot2021}\right)\)
Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).
Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\), \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\), \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\), \(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)
\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)
Tính:
\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
Thực hiện phép tính
\(A=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{1+2+3+..+2006}\right)\)