Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)

\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)

Sửa đề: \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2019\cdot2021}\right)\)

Ta có: \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2019\cdot2021}\right)\)

\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{2020^2-1}\right)\)

\(=\frac{2^2-1+1}{2^2-1}\cdot\frac{3^2-1+1}{3^2-1}\cdot\ldots\cdot\frac{2020^2-1+1}{2020^2-1}\)

\(=\frac{2^2}{2^2-1}\cdot\frac{3^2}{3^2-1}\cdot\ldots\cdot\frac{2020^2}{2020^2-1}\)

\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\ldots\cdot\frac{2020\cdot2020}{2019\cdot2021}\)

\(=\frac{2\cdot3\cdot\ldots\cdot2020}{1\cdot2\cdot\ldots\cdot2019}\cdot\frac{2\cdot3\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2021}=\frac{2020}{1}\cdot\frac{2}{2021}=\frac{4040}{2021}\)

a, ( - 7 ) x ( - 8 ) = 56

b, 3 / 7 x ( - 2 / 9 ) = - 6 / 63 = - 2 / 21

c, - 7 / 6 : 7 / 12 = - 7 / 6 x 12 / 7 = - 12 / 6 = -2

d, - 1 / 2 x 3 / 4 + 1 / 4 x ( - 1 / 2 ) = - 1 / 2 x ( 3 / 4 + 1 / 4 ) = - 1 / 2 x 1 = - 1 / 2

S=(1/1.3+1/3.5+.....+1/7.9)+(1/2.4+1/4.8+1/8.10)

2S=1/2.(1-1/3+1/5-1/5+....+1/7-1/9)+(1/2-1/4+1/4-1/8+1/8-1/10)

2S=1/2.(1-1/9)+(1/2-1/10)

2S=1/2.(8/9+2/5)

bài này dễ mà bạn bạn chỉ cần đổi ra rồi tính bình thường là đc mà

Bài 1:

\(A=2^2\cdot3-4\\ =4\cdot3-4\\ =4\cdot\left(3-1\right)\\ =4\cdot2\\ =8\\ B=16-2^3\cdot2\\ =16-16\\ =0\\ C=4^2-4\cdot2\\ =4\cdot\left(4-2\right)\\ =4\cdot2\\ =8\\ D=3^3-3\cdot3^2\\ =3^3-3^3\\ =0\)