Mấy bạn trả lời cho mk đi mà .

Cứu tôi vs , tôi sắp chết nếu như ko ai giải cho tôi câu này

99/50 mk nhanh nhất k mk nha

\(\frac{22}{1\cdot3}\cdot\frac{32}{2\cdot4}\cdot\frac{42}{3\cdot5}\cdot...\cdot\frac{992}{98\cdot100}\)

Mk vt lại đề nè bn xem có đúng ko

Tính: 2² phần 1.3 . 3² phần 2.4 . 4² phần 3.5 ...... 99² phần 98.100 = 2² phần 1.3 . 3² phần 2.4 . 4² phần 3.5 ...... 99² phần 98.100

Mình vt lộn đề mình viết lại và giải nè

\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{99^2}{98\cdot100}\)

\(\Rightarrow\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{99\cdot99}{98\cdot100}\)

\(\Rightarrow\frac{2\cdot3\cdot4\cdot...\cdot99}{1\cdot2\cdot3\cdot...\cdot98}\cdot\frac{2\cdot3\cdot4\cdot....\cdot99}{3\cdot4\cdot5\cdot...\cdot100}\)

\(\Rightarrow\frac{99}{1}\cdot\frac{2}{100}=\frac{99}{1}\cdot\frac{1}{50}=\frac{99}{50}\)

\(B=\left(1+\frac{1}{1.3}\right)+\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...+\left(1+\frac{1}{98.100}\right)\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\right)\)( 98 số 1 ở tồng đầu tiên)

\(=98+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.101}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

\(=98+\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{3}{97.101}\right)+\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(=98+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{98}-\frac{1}{100}\right)\)\(=98+\frac{1}{2}.\left(1-\frac{1}{101}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=98+\frac{1}{2}.\frac{100}{101}+\frac{1}{2}.\frac{49}{100}\)

\(=98+\frac{51}{101}+\frac{49}{200}\)

Suy ra phàn nguyên của B là 98.

Vậy phân fnguyên của B là 98.

mình nhầm. bạn thay các chỗ có "97.101" thành "99.101" nhé!

Xét : \(\frac{x^2}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{x^2-1}\) 

=> \(\left[\frac{x^2}{x^2-1}\right]=1\) vì \(0< \frac{1}{x^2-1}< 1\)

Do đó : \(\left[D\right]=1.98=98\) 

\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A

\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)

\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)

Dễ thấy phần nguyên của B là 98

Vậy phần nguyên của B là 98

B=1.3+2.4+3.5+...+97.99+98.100B=1.3+2.4+3.5+...+97.99+98.100

B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)

B=1.2+1+2.3+2+....+97.98+97+98.99+98B=1.2+1+2.3+2+....+97.98+97+98.99+98

B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)

B=98.99.1003+98.992B=98.99.1003+98.992

B=323400+4851=328251B=323400+4851=328251 

Số đó=1.3 + 2.4 + 3.5 +....+ 98.100 
= 1(2+1) + 2.(3+1) + 3.(4+1) +...+ 98(99+1)
= 1.2 + 1 + 2.3 + 2 + 3.4 + 3+....+ 98.99 +98
= (1.2 + 2.3 + 3.4+....98.99) + (1+2+3+....+98)
=323400 + 4851=328251

Đề sai nhá dãy số lẻ ko thể kết thúc bằng số chẵn đc  : 

Đề này nhá \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}\)

\(\Rightarrow A=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{101}\right)\)

\(\Rightarrow A=2.\frac{100}{101}=\frac{200}{101}\)

\(A=\frac{4}{1.3}+\frac{4}{3.5}+....+\frac{4}{98.100}\)

\(A=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+.........+\frac{2}{98.100}\right)\)

\(A=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{98}-\frac{1}{100}\right)\)

\(A=2.\left(1-\frac{1}{100}\right)\)

\(A=2.\frac{99}{100}\)

\(A=\frac{99}{50}\)

= 1.( 2 + 1 ) + 2 ( 3 + 1 ) + 3( 4 + 1 ) + ..... + 98( 99 + 1 )

 = 1 . 2 + 1 + 2 . 3 + 2 + 3 . 4 + 3 + ...... 98 . 99 + 98

= ( 1 . 2 + 2 . 3 + 3 . 4 + ..... + 98 . 99 ) + ( 1 + 2 + 3 + ..... + 98 )

= 323400 + 4851 

= 328351

bn ủng hộ vũ khánh linh nha

A = 1.( 2 + 1 ) + 2.( 3 + 1 ) + 3.( 4 + 1 ) + .... + 98.( 99 + 1 )

A = 1.2 + 1 + 2.3 + 2 + 3.4 + 3 + .... + 98.99 + 98

A = ( 1.2 + 2.3 + 3.4 + .... + 98.99 ) + ( 1 + 2 + 3 + ... + 98 )

Đặt B = 1.2 + 2.3 + 3.4 + .... + 98.99

      C = 1 + 2 + 3 + .... + 98

=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + .... + 98.99.3

          = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ..... + 98.99.( 100 - 97 )

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .... + 98.99.100 - 97.98.99

          = 98.99.100

=> B = ( 98.99.100 ) : 3 = 323400

=> C =  1 + 2 + 3 + .... + 98

     C = 98 + 97 + .... + 2 + 1

=> 2C = 99 + 99 + .... + 99 + 99

=> 2C = 99.98 = 9702

=> C = 4851

\(\Rightarrow A=323400+4851=328251\)