\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+...+\frac{100^2}{99.101}\\ =\frac{2.2}{1.3}+\frac{3.3}{2.4}+...+\frac{100.100}{99.101}\\ =\frac{2.}{1.}\frac{3.}{2.}\frac{...}{...}\frac{100}{99}+\frac{2.}{3.}\frac{3.}{4.}\frac{...}{...}\frac{100}{101}\\ =\frac{100}{1}+\frac{2}{101}\\ =\frac{10100}{101}+\frac{2}{101}\\ =\frac{10102}{101}\)

\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{100^2}{99.101}\)

\(=\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+...+\frac{100.100}{99.101}\)

\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}\)

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

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

=200/101 nha bạn

\(\frac{200}{101}\)nha! k nhé

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.,,\frac{50^2}{49.51}\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.,,\frac{50.50}{49.51}\)

=\(\frac{\left(2.3.4...50\right).\left(2.3.4...50\right)}{\left(1.2.3....49\right).\left(3.4.5....51\right)}\)

=\(\frac{50.2}{1.51}\)

=\(\frac{100}{51}\)

\(=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4...50}{3.4.5...51}=50.\frac{2}{51}=\frac{100}{51}\)

A=2^2/1.3+3^2/2.4+4^2/3.5+....+99^2/98.100

A=2^2/(2-1)(2+1)+3^2/(3-1)(3+1)+4^2/(4-1)(4+1)+...+99^2/(99-1)(99+1)

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

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

A=1+1/1.3+1+1/2.4+1+1/3.5+...+1+1/98.100

A=(1+1+1+....+1)+(1/1.3+1/2.4+...+1/98.100) (1)

Ta có:

Đặt B=(1+1+1+...+1)=98[vì (99-2):1+1=98 số] (2)

Đặt C=1/1.3+1/2.4+1/3.5+...+1/98.100

=>C=1/2.(1-1/3)+1/2.(1/2-1/4)+1/2.(1/3-1/5)+...+1/2.(1/98-1/100)

=>C=1/2.(1-1/3+1/2-1/4+1/3-1/5+...+1/97-1/99+1/98-1/100)

=>C=1/2.(1+1/2-1/99-1/100)

=>C=1/2.(3/2-1/99.100) (3)

Thay (2),(3) vào(1), được:

A=98+1/2.(3/2-1/99.100)

????????????????????????????????

\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)

\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)

\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)

=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????

ai choi freefire thi kb voi minh

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

\(=1-\frac{1}{101}\)

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

2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2\times\frac{502}{1005}\)

\(=\frac{1004}{1005}\)

tự làm tiếp nhé

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

   = \(1-\frac{1}{101}\) = \(\frac{100}{101}\)

2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

   = \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

   = \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)

a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

b)\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+..+\frac{4}{2008.2010}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1004}-\frac{1}{1005}=1-\frac{1}{1005}=\frac{1004}{1005}\)

c)\(\frac{1}{2}.\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{9800}\right)=\frac{1}{4}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{400}\)

a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101 

=1-1/101 

=100/101 

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5 

=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5 

=(1-1/101).2,5

=100/101.2,5 

=250/101 

dấu / là phần nhé. bạn có thể xem bài có dấu phần ở : Câu hỏi của Nguyễn Thị Hoài Anh 

A)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

=1-\(\frac{1}{101}\)

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

B) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{1}{99.101}\)

=5.(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))

=5.\(\frac{2}{2}.\)(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))

=5.\(\frac{1}{2}\).(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\))

=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))

=\(\frac{5}{2}.\frac{100}{101}=\frac{250}{100}\)

Chúc bạn học tốt

*\(\frac{x}{200}\)=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\)....\(\frac{99^2}{99.100}\)

=>\(\frac{x}{200}\)=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{99}{100}\)

=>\(\frac{x}{200}\)=\(\frac{1}{100}\)

=>100x=200

=>x=2