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

sai đề 

sai đề là cái chắc

sai đề hả bạn????

...

= 1/2-1/3+1/3-1/4+...+ 1/19-1/20

= 1/2-1/20

=9/20

có phải như thế này ko bn

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)

A = \(\frac{9}{20}\)

\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

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

B = \(-\frac{99}{100}\)

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

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

\(A=\frac{1}{2}-\frac{1}{20}\)

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

Tương tự với ý B

có: 1/1.2= 1/1-1/2

1/2.3=1/2-1/3

...

1/8.9= 1/8-1/9

1/9.10=1/9-1/10

suy ra B=1/1-1/2+1/2-1/3+...+1/9-1/10

=1/1-1/10 (vì bạn thấy trừ 1/2 rồi cộng 1/2, trừ 1/3 cộng 1/3,... thì khử đi)

=9/10.

vậy B=9/10

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

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

1.50+2.49+3.48+...+49.2+50.1=

= (1.50+2.50+3.50+...+50.1)-(1.2+2.3+3.4+...+49.50)

= (2500+50).50:2-41650

= 63750-41650=22100

2, 

A = 1.2 + 2.3 + 3.4 + ... + 2011.2012

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012

3A = 2011.2012.2013

A = 2011.2012.2013 : 3 

A = 2714954572

1)A=22100

2)B=60236

= -101/100

\(B=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\\ =-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\\ =-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\\ =-\left(1-\frac{1}{100}\right)=\frac{-99}{100}\)

<br class="Apple-interchange-newline"><div></div>B=−11.2 −12.3 −13.4 −...−198.99 −199.100  \\ =−(11.2 +12.3 +13.4 +...+198.99 +199.100 ) \\ =−(1−12 +12 −13 +13 −14 +...+198 −199 +199 −1100 ) \\ =−(1−1100 )=−99100 

Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

=>\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

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

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

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

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

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

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

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

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

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

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

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

Hình như bn viết sai đề,là 1/x.(x+1) chứ

ukm mik xin lỗi mik viết sai đề đó

Đề sai nhé phải là x(x+1)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\Leftrightarrow\frac{x}{x+1}=\frac{2015}{2016}\Rightarrow x=2015\)

Vậy \(x=2015\)

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3