Ta có: A=1.2+2.3+...+198.199+199.200 
=>3A=1.2.3+2.3.3+...+198.199.3 
+199.200.3 
=>3A=1.2.3+2.3(4-1)+...+ 
198.199(200-197)+199.200(201-198) 
=>3A=1.2.3+2.3.4-1.2.3+...+198.199.200 
-197.198.199+199.200.201-198.199.200 
=>3A=199.200.201 
=>A=199.200.67

A=39800.67

A=2666600

lấy máy tính mà tính

các bạn troll mình à

3A =1.2.3 +2.3.(4-1) +3.4.(5-2) +4.5.(6-3)....+199.200.(201 -198)

    = 1.2.3+2.3.4 -1.2.3 +3.4.5- 2.3.4 + 4.5.6 - 3.4.5 +......+ 199.200.201 -198.199.200

 3A =199.200.201 

A=199.200.67 =254600

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

=\(1-\dfrac{1}{5}\)

=\(\dfrac{4}{5}\)

1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5

 = 1/1 - 1/5

= 4/5

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{200.201}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)

=\(\frac{1}{2}-\frac{1}{201}\)

=\(\frac{199}{402}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{200.201}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{200}-\frac{1}{201}\)

\(=\frac{1}{2}-\frac{1}{201}=\frac{199}{402}\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{99.100}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

= \(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(A=1-\frac{1}{200}\)

\(A=\frac{199}{200}\)

\(=1-\frac{1}{200}=\frac{199}{200}\)

\(A=\frac{1}{1.2}+............+\frac{1}{199.200}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...........+\frac{1}{199}-\frac{1}{200}\)

\(=1-\frac{1}{200}\)

\(=\frac{199}{200}\)

\(S=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.......+\frac{1}{49\cdot50}\)

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{49}+\frac{1}{50}\)

\(S=\frac{1}{2}-\frac{1}{50}\)

\(S=\frac{25}{50}-\frac{1}{50}\)

\(S=\frac{24}{50}=\frac{12}{25}\)

ai k mh mh k lại

k cho mh nha

S=1/2.3+1/3.4+1/4.5+....+1/49.50

=\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...........+\frac{1}{49x50}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{49}-\frac{1}{50}\)

=\(\frac{1}{2}-\frac{1}{50}\)

=\(\frac{24}{50}\) mình cũng ko chắc đúng nhưng đây là cách giải của mình

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{12}{25}\)

Hh

P = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  ... + 1/8 - 1/9

= 1/2 - 1/9

= 7/18

P = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  ... + 1/8 - 1/9

= 1/2 - 1/9

= 7/18.

A= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6

A= 1/2 - 1/3+ 1/3-1/4 + 1/4-1/5+ 1/5-1/6

A= 1/2- 1/6

A= 1/3

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{6}\)

\(\Rightarrow A=\frac{1}{3}\)

Wa cảm ơn các bạn nhiều nha