\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

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

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

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

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

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)

\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\frac{98}{99}\)

\(N=\frac{49}{33}\)

\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{97\cdot99}=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]\)

\(=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{5}{2}\left[1-\frac{1}{99}\right]\)

\(=\frac{5}{2}\cdot\frac{98}{99}=\frac{245}{99}\)

\(=\frac{5}{2}\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{5}{2}\times\frac{98}{99}\)

\(=\frac{245}{99}\)

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

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

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

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

\(A=\frac{3}{1.3}+\frac{3}{3.5}+.....+\frac{3}{19.21}\)

\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{19.21}\right)\)

\(A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{19}-\frac{1}{21}\right)\)

\(A=\frac{3}{2}.\left(1-\frac{1}{21}\right)\)

\(A=\frac{3}{2}.\frac{20}{21}\)

\(A=\frac{10}{7}\)

Ta có:

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{19.21}\)

\(\Rightarrow A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)

\(\Rightarrow A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{21}\right)=\frac{2}{3}.\frac{20}{21}=\frac{40}{63}\)

6B=1.3.(5+1)+3.5.(7-1)+...+19.21(23-17)

6B=1.3.5+1.3+3.5.7-1.3.5+...+19.21.23-17.19.21

6B=3+19.21.23

6B=9180

 B=9180/6

   B=1530

tick nhé ko tick mai đến lớp tao phang

lm ah vũ

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\) 

\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)

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

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

= \(\dfrac{5}{2}-\dfrac{100}{101}\)

= \(\dfrac{305}{202}\)

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

A= 2/1x3 + 2/3x5 + 2/5x7 +... + 2/2003x2005

A= 1 - 1/3 +1/3 - 1/5  + 1/5 - 1/7 + ... + 1/2003 + 1/2005

A= 1 - 1/2005

A= 2004/2005

B= 2006/2005

suy ra A < B

Ta co :A=2004/2005 vay thi A<B roi