\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)

     \(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

       \(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)

      \(=5.\left(1-\frac{1}{100}\right)\)

   \(=5.\frac{99}{100}\)

      \(=\frac{99}{20}\)

5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100

= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )

= 5 . ( 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 )

= 5 . ( 1 - 1/100 )

= 5 . 99/100

= 99/20

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow A< 1\)

b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)

\(\Rightarrow2B< 1\)

\(\Rightarrow B< \frac{1}{2}\)

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

A = \(\frac{1}{1}\)-\(\frac{1}{100}\)

ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây 

A = \(\frac{99}{100}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(A=1-\frac{1}{100}\)

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

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

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

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

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

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Số x cần tìm là : 99

Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

=> x + 1 = 100

=> x = 99

\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+...+\frac{181}{9.10}\)

=\(\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+...+\frac{180+1}{90}\)

=\(2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+...+2+\frac{1}{9.10}\)

=\(18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

=\(9-\frac{1}{10}\)

=\(\frac{189}{10}\)

1.

a. \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

\(=5.\left(1-\frac{1}{100}\right)\)

\(=5.\frac{99}{100}\)

\(=\frac{99}{20}\)

b. \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

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

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

\(=\frac{4}{2}.\left(1-\frac{1}{101}\right)\)

\(=2.\frac{100}{101}\)

\(=\frac{200}{101}\)

b) \(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot...\cdot\frac{100^2}{100\cdot101}=\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{1\cdot2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{2\cdot3\cdot4\cdot...\cdot101}=1\cdot\frac{1}{101}=\frac{1}{101}\)

a không biết

câu b mình thiếu, là \(\frac{100^2}{100.101}\)nhé

Gần đúng được không

\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)cách làm \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) tách hết ra bạn thấy cái giữa tự triệt tiêu nhau

\(B=\frac{5}{1.3}+...+\frac{5}{23.25}\) { nếu đúng là \(\frac{5}{23.5}\)  thì làm đến \(\frac{5}{21.23}\) rồi cộng lẻ cái cuối

\(\frac{2B}{5}=\frac{2}{1.3}+...+\frac{1}{23.25}=1-\frac{1}{25}\) cách làm giống (a)

\(\frac{2}{5}B=1-\frac{1}{25}=\frac{24}{25}\Rightarrow B=\frac{24}{25}.\frac{5}{2}=\frac{12}{5}\)

A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

=\(\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{19}-\frac{1}{20}\right)\)

=\(\frac{1}{2}-\frac{1}{20}=\frac{10-1}{20}=\frac{9}{20}\)

a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)

\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)

\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)

Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)

\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)

\(2A=\frac{12}{3}-\frac{12}{99}\)

\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)