a)

\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)

\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)

b)

\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)

\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)

\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)

\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)

\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)

c)

\(C=\frac{10}{7,12}+\frac{10}{12.17}+...+\frac{10}{502.507}\)

\(\Rightarrow \frac{C}{2}=\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\)

\(\frac{C}{2}=\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{507-502}{502.507}\)

\(\frac{C}{2}=\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{502}-\frac{1}{507}\)

\(\frac{C}{2}=\frac{1}{7}-\frac{1}{507}=\frac{500}{3549}\)

\(\Rightarrow C=\frac{1000}{3549}\)

a,A=1/5-1/8+1/8-1/11+...+1/2006-1/2009=1/5-1/2009=2004/10045

b,B=1/4x(4/6x10+4/10x14+...+4/402x406)

=1/4x(1/6-1/10+1/10-1/14+...+1/402-1/406)

=1/4x(1/6-1/406)

=1/4x100/609=25/609

c,C=2x(5/7x12+5/12x17+...+5/502x507)

=2x(1/7-1/12+1/12-1/17+...+1/502-1/507)

=2x(1/7-1/507)

=2x500/3549

=1000/3549

Xin lỗi vì ko viết được rõ ràng.Mong bạn thông cảm. Chúc bạn học tốt.

\(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{2006\times2009}\)

\(=\frac{1}{3}\left(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{2006\times2009}\right)\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\right)\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{2009}\right)\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{2009}\right)\)

\(=\frac{1}{3}\left(\frac{2009}{10045}-\frac{5}{10045}\right)\)

\(=\frac{1}{3}.\frac{2004}{10045}=\frac{2004}{30135}\)

A= 1/4 ( 1/2x6 +1/6x10 +.............+1/46x50)

A= 1/4 ( 1/2 - 1/6 + 1/6 - 1/10 +.......... + 1/46 - 1/50 )

A= 1/4 ( 1/2 - 1/50 )

A= 1/4 x 12/25

A= 3/25

3/25 nha

thanks nha

Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)

\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\) (**)

Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)

\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm

b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)

\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)

\(3VT=1-\dfrac{1}{64}< 1\)

\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)

ngu như con lợn

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )

\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)

\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)

\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)

Rồi làm tương tự cân b nha!

\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)

\(+\dfrac{1}{57}-\dfrac{1}{87}\)

\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)

Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:

a) \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{1}{1.2}\\\dfrac{1}{6}=\dfrac{1}{2.3}\\\dfrac{1}{12}=\dfrac{1}{3.4}\\...\end{matrix}\right.\)

Vậy số thứ 100 của dãy là: \(\dfrac{1}{100.101}=\dfrac{1}{10100}\)

Tổng: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

b) \(\left\{{}\begin{matrix}\dfrac{1}{6}=\dfrac{1}{\left(5.0+1\right)\left(5.1+1\right)}\\\dfrac{1}{66}=\dfrac{1}{\left(5.1+1\right)\left(5.2+1\right)}\\\dfrac{1}{176}=\dfrac{1}{\left(5.2+1\right)\left(5.3+1\right)}\\...\end{matrix}\right.\)

Vậy số thứ 100 của dãy là: \(\dfrac{1}{\left(5.99+1\right)\left(5.100+1\right)}=\dfrac{1}{248496}\)

Tổng: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{496.501}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{496.501}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)

\(=\dfrac{1}{5}.\dfrac{500}{501}\)

\(=\dfrac{100}{501}\)

Bài 2: Tính:

a) \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)

\(A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

\(A=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

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

Bài 2 :

a, Xét tử số : Đặt B = \(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}\)

Số số hạng của tử số là : ( 99 - 1 ) : 2 + 1 = 50 ( số )

=> Tử số có 50 phân số

Ta có : \(B=\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+\left(\dfrac{1}{5}+\dfrac{1}{95}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)\)

\(=\left(\dfrac{99}{99}+\dfrac{1}{99}\right)+\left(\dfrac{97}{3.97}+\dfrac{3}{3.97}\right)+\left(\dfrac{95}{5.95}+\dfrac{5}{5.95}\right)+...+\left(\dfrac{51}{49.51}+\dfrac{49}{49.51}\right)\)

\(=\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}\)

Xét mẫu số : Đặt C = \(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}\)

\(=\left(\dfrac{1}{1.99}+\dfrac{1}{99.1}\right)+\left(\dfrac{1}{3.97}+\dfrac{1}{97.3}\right)+...+\left(\dfrac{1}{49.51}+\dfrac{1}{51.49}\right)\)

\(=2.\dfrac{1}{1.99}+2.\dfrac{1}{3.97}+...+2.\dfrac{1}{49.51}\)

\(=2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)\)

Thay B và C vào A ta có :

\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

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

Vậy A = 50

b, Xét mẫu số : Đặt C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)

\(=\dfrac{100-1}{1}+\dfrac{100-2}{2}+\dfrac{100-3}{3}+...+\dfrac{100-99}{99}\)

\(=100-1+\dfrac{100}{2}-1+\dfrac{100}{3}-1+...+\dfrac{100}{99}-1\)

\(=\left(100+\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}\right)-\left(1+1+...+1\right)\)

Đặt D = 1 + 1 + ... + 1

Số số hạng của tổng D là : ( 99 - 1 ) : 1 + 1 = 99 ( số hạng )

\(\Rightarrow D=1.99=99\)

Thay D = 99 ta có :

\(C=100\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-99\)

\(=100+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-99\)

\(=\left(100-99\right)+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)\)

\(=1+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)\)

\(=\dfrac{100}{100}+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)

Thay vào đề bài , ta có :

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

Vậy \(B=\dfrac{1}{100}\)