a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)

A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)

A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)

A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)

A = \(34+\frac{34}{103}=\frac{3536}{103}\)

bạn làm hộ mik câu B với

\(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{49.51}{99.101}\)

\(\Rightarrow Q=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(50-1\right)\left(50+1\right)}{\left(50.2-1\right)\left(50.2+1\right)}\)

\(\Rightarrow Q=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...+\)\(\frac{1}{4}-\frac{3}{8}\left(\frac{1}{50.2-1}-\frac{1}{50.2+1}\right)\)

\(\Rightarrow Q=49.\frac{1}{4}-49.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+...+\frac{1}{50.2-1}-\frac{1}{50.2+1}\right)\)

\(\Rightarrow Q=\frac{49}{4}-\frac{147}{8}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow Q=\frac{49}{4}-\frac{147}{8}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(\Rightarrow Q=\frac{49}{4}-\frac{147}{8}.\frac{98}{303}\)

\(\Rightarrow Q=\frac{49}{4}-\frac{2401}{404}\)

\(\Rightarrow Q=\frac{637}{101}\)

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

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

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

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

\(b,=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)

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

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

a,\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=1.\frac{99}{100}=\frac{99}{100}\)

ồ nhầm đề

Lồ

hơi khó đó tick mình nha Hoàng Thu Hà

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

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

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

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

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

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

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

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

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

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

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

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

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

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

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

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

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

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

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

\(\frac{x}{2^2}\)+\(\frac{x}{2^3}\) +\(\frac{x}{2^4}\) =\(\frac{x}{3^2}\) +\(\frac{x}{3^3}\) +\(\frac{x}{3^4}\) là x =

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

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

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

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

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

\(=\frac{250}{101}\)

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

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