Dãy số trên có số số hạng là: (khoảng cách mỗi số là $1$ đơn vị)

$(2020-1):1+1=2020$(số hạng)

Tổng của dãy số trên là:

$(2020+1)\times2020:2=2041210$

1+2+3+...+2020=\(\dfrac{\left(2020-1\right):1+1\cdot\left(1+2020\right)}{2}\)=2041210.

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

M = 5 + 5³ + 5⁵ + ... + 5⁴⁷ + 5⁴⁹

5²M = 5²(5 + 5³ + 5⁵ + ... + 5⁴⁷ + 5⁴⁹)

25M = 5³ + 5⁵ + 5⁷ + ... + 5⁴⁹ + 5⁵¹

25M - M = ( 5³ + 5⁵ + 5⁷ + ... + 5⁴⁹ + 5⁵¹) - (5 + 5³ + 5⁵ + ... + 5⁴⁷ + 5⁴⁹)

24M = 5⁵¹ - 5

M = \(\frac{5^{51}-5}{24}\)

Còn mấy câu kia bạn biết ko?

\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{97}{98}\cdot\dfrac{98}{99}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\)

\(=\dfrac{1}{99}\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{100}{2}=50\)

Mik cần gấp mn ko giúp đc thì thôi ạ

Mình trình bày trong hình nhé ^^ Bn tham khảo ạ

\(3\dfrac{2}{5}\cdot1\dfrac{4}{7}=\dfrac{17}{5}\cdot\dfrac{11}{7}=\dfrac{187}{35}\)

1) I ( not play ) .am not playing.. football now . I'm tired

2) Shh ! they ( study ) .are studing.. in the library 

3) Vy and Phong ( not do ) .doesn't.. their homework

4) Now I ( have ) .am having.. an English leson with Mr Lee

5) Phong  , Vy and Duy ( ride ) .are riding.. their bicycles to school now

1) I am not playing football now . I'm tired

2) Shh ! they are studying in the library 

3) Vy and Phong aren't doing their homework

4) Now I am having an English leson with Mr Lee

5) Phong  , Vy and Duy are riding their bicycles to school now

các bạn ơi giúp mình với !

\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)

\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)

\(=\dfrac{75}{101}\)

\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)

\(=\dfrac{100}{2}=50\)

Tính nhanh:

a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)

= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{3}{4}\times\dfrac{100}{101}\)

= \(\dfrac{75}{101}\)

b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)

\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)

\(=\dfrac{100}{2}\)

\(=50\)