Giải:

\(\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{6}{352}\) 

\(=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\) 

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}\) 

\(=\dfrac{1}{1}-\dfrac{1}{22}\) 

\(=\dfrac{21}{22}\)

\(\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{6}{352}\\ =\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{3}{176}\\ =\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{1}{22}\\ =\dfrac{21}{28}+\dfrac{3}{28}+\dfrac{7}{154}+\dfrac{8}{154}\\ =\dfrac{6}{7}+\dfrac{15}{154}\\ =\dfrac{21}{22}\)

\(\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}+\dfrac{6}{352}=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}=1-\dfrac{1}{22}=\dfrac{21}{22}\)

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

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 \(B=\dfrac{1}{100}\)

(3/7+6/7-2/7):1/5

=1:1/5

=5

=\(\left(\dfrac{3}{7}+\dfrac{6}{7}-\dfrac{2}{7}\right):\dfrac{1}{5}=1:\dfrac{1}{5}=5\)

Mình nghĩ đề là : 2/8 sẽ hay hơn.

\(B=\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{2}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)

\(=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{2}{8}+\dfrac{6}{8}\right)\)

\(=6+1+1=8\)

\(B=\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{3}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)

\(B=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{3}{8}+\dfrac{6}{8}\right)\)

\(B=6+1+1,125\)

\(B=8,125\)

 = \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)

 = \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

 = \(\dfrac{5}{2}.\dfrac{100}{101}\)

 = \(\dfrac{250}{101}\)

\(=\dfrac{5\left(5+2\right)}{11.7}+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=1\)

Ta có:

= 5/11 .( 5/7+ 2/7 ) + 6/11

= 5/11 . 1 + 6/11

= 11/ 11

= 1

#học tốt ạ :3

`6/11+2/5+16/11+19/13+3/5+7/13`

`=(6/11+16/11)+(2/5+3/5)+(19/13+7/13)`

`=22/11+5/5+26/13`

`=2+1+2=5`

\(=\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{19}{13}+\dfrac{7}{13}\right)\)

\(=2+1+2\)

\(=5\)