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

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

= 

ta có 

\(C=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{4.3}+..+\frac{100-99}{99.100}\)

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

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

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

\(=1-\frac{99}{100}\)

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

Mọi thứ trong thế giới này chỉ là một trò chơi và chúng ta chỉ là những con tốt...

ịt địt đị

câu a) (a^2+2a+a+2)(a+3)-(a^2+a)(a+2)= (3a+3)(a+2)

suy ra: a^3+3x^2+2a^2+6a+a^2+3a+2a+6-a^3-2x^2-a^2-2a= 3a^2+6a+3a+6

3a^2+9a+6=3a^2+9a+6

câu b) 

^ là gì vậy bạn

Toán tiểu học: dang phân số có tử số là hiệu của hai thừa số ở mẫu

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

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

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

Gọi tổng trên là A

A = 1/1.2 + 1/2.3 +......+ 1/99.100

A = 1 - 1/2 + 1/2 - 1/3 +.......+1/99 - 1/100

A = 1 - 1/100

A = 99/100

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

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

= \(1-\frac{1}{50}=\frac{49}{50}\)

B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)

= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)

= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

= \(\frac{4}{13}\)

C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\) 

= \(\frac{9}{38}\)

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

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

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

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

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

=1/1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100

=1-1/100

=99/100

=1−1/2+1/2−1/3+1/3−1/4+...+1/99−1/100
=1 − 1/100 = 99/100

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100