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\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).........................\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\left(\dfrac{1}{4}-\dfrac{4}{4}\right)................\left(\dfrac{1}{99}-\dfrac{99}{99}\right)\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)
\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)
\(A=\dfrac{-1}{100}\)
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
=-1+(4-7)+(10-13)+(16-19)+...+(94-97)+(100-103)
=-1+(-3)+(-3)+(-3)+...+(-3)+(-3)(17 cặp)
=-1+(-3)*17
=-1+(-51)
=-52
A = 1 -2 + 22 - 23 + 24 - 25 + ... + 22008
2A = 2 .(1 -2 + 22 -23 +24 -25 +...+ 22008)
2A = 2 - 22 + 23 - 2 4+ 25 - 26 + ... + 22009
2A-A= ( 2 -22+23-24+25-26+...+22009) - (1-2+22-23+24-25+...+22008)
A = (2-2) + (-22+22) + ..... + ( 22008 - 22008) + ( 22009 - 1)
A = 0 + 0 + ... + 0 + 22009 -1
A = 22009 - 1
`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`
`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`
`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`
`=(1-1/7).[130-4-125]/3`
`=6/7 . 1/3 = 2/7`
____________________________________________________
`8/9+1/9 . 2/9+1/9 . 7/9`
`=8/9+1/9.(2/9+7/9)`
`=8/9+1/9 . 9/9`
`=8/9+1/9=9/9=1`
a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)
\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)
\(=-\dfrac{5}{24}\)
b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)
\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)
\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{7}\)
\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)
\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=1\)
ta có :
1 đến 100 =
có 100 số hạng
tổng : ( 100 + 1 ) x 100 : 2 = 5050
công thức : ta lấy số lớn nhất trừ số bé nhất trong dãy số trong ngoặc chia số số hạng rồi chia 2
12 + 22 + 32 ........ + 102 =
12 = 1 x 1 = 1 62 = 6 x 6 = 36
22 = 2 x 2 = 4 72 = 7 x 7 = 49
32 = 3 x 3 = 9 .......
42 = 4 x 4 = 16 tổng dãy số là : 1 + 9 + 4 + 16 + 25 + 36 .... = 395
52 = 5 x 5 = 25
65 x 111 x 13 x 15 x 37 = 52056225
Đ/s : ta lấy 3 kết quả nhân với nhau rồi xem có thể xử dụng lũy thừa để rút gọn hay không