xin lỗi mình mới học lớp 5 thôi

Có cần giải tóm tắt không

chám là cái gì

\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)

\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)

\(A=\frac{4.1.4.1}{3.2.15.6}\)

\(A=\frac{4}{135}\)

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)

\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)

\(=\frac{5}{1}.\frac{2}{6}\)

\(=\frac{5}{1}.\frac{1}{3}\)

\(=\frac{5}{3}\)

Đọc kỹ lại đề đi Trần Cao Vỹ Lượng

C=2.2.3.3.4.4.5.5/1.3.2.4.3.5.4.6

C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)

C=2.5/6

C=5/3

C=2^2/1.3.3^2/2.4.5^2/3.5.4^2/4.6

C= 2.2.3.3.5.5.4.4/1.3.2.4.3.5.4.6

C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)

C=5.2/6

C=5/3

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{5^2}{3.5}.\frac{4^2}{4.6}\)

\(=\frac{2^2.3^2.5^2.4^2}{1.3.2.4.3.5.4.6}\)

\(=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}\)

\(=\frac{2.5}{6}\)

\(=\frac{10}{6}=\frac{5}{2}\)

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

=4/3.9/8.16/15.25/4096(CÁI NÀY MÌNH BIẾN NÓ VỀ PHÂN SỐ NHA)

=5/512 (BẠN THỰC HIỆN PHÉP TÍNH TỪ TRÁI SANG PHẢI NHÉ)

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4^6}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{4096}=\frac{4.9.16.25}{3.8.15.4096}=\frac{3.5}{2.3.256}=\frac{5}{2.256}=\frac{5}{512}\)

# Học tốt.!

=)))

Jang Ha Na làm sai nhé!!!

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

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

  \(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(=\frac{4}{9}-\frac{1}{5}\)

\(=\frac{11}{45}\)