\(\frac{5}{1.4}+\frac{5}{4.7}+................+\frac{5}{100.103}\)

\(\frac{1}{3}.\left(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+..............+\frac{5}{100}-\frac{5}{103}\right)\)

\(\frac{1}{3}.\left(5-\frac{5}{103}\right)\)

\(\frac{1}{3}.\left(\frac{510}{103}\right)=\frac{170}{103}\)

\(=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{....}{....}\)

xin lỗi mình ko có máy tính nên ko tính đc phép tính cuối

S = \(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+.......+\frac{5}{100}-\frac{5}{103}\)

S = \(5-\frac{5}{103}\)

S = \(\frac{510}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

=5/3.(3/1.4+3/4.7+3/7.10+...+3/100.103)

=5/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)

=5/3.(1-1/103)=5/3.102/103=170/103

                                       đáp số : 170/103

a cong tru loan nen ko hieu

b

A=5/1.4+5/4.7+..5/100.103

3/5.A=3/1.4+3/4.7+..+3/100.103

=1/1-1/4+1/4-1/7+...+1/100-1/103

=1-1/103=102/103

A=(5.102)/(3.103)=5.34/103

=\(\frac{5}{3}\cdot\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\right)\)

=\(\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

=\(\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)\)

=\(\frac{5}{3}\cdot\frac{102}{103}\)=\(\frac{170}{103}\)

Vậy D=\(\frac{170}{103}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)

\(3B=5\left(1-\frac{1}{103}\right)\)

\(3B=5.\frac{102}{103}\)

\(3B=\frac{510}{103}\)

\(\Rightarrow B=\frac{170}{103}\)

Ta có:

B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}+.....+\frac{5}{100.103}\)

B=\(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{100.103}\right)\)

B=\(\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\right)\)

B=\(\frac{5}{3}\left(1-\frac{1}{103}\right)\)

B=\(\frac{5}{3}.\frac{102}{103}\)

B=\(\frac{170}{103}\)

Vậy B=\(\frac{170}{103}\)

nhớ k

Trả lời

\(B=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\)

\(\frac{3}{5}B=\frac{3}{5}\left(\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\right)\)

\(\frac{3}{5}B=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{...3}{100\cdot103}\)

\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

\(\frac{3}{5}B=1-\frac{1}{103}\)

\(\frac{3}{5}B=\frac{102}{103}\)

\(B=\frac{102}{103}:\frac{3}{5}\)

\(B=\frac{170}{103}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)

\(3B=15\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(3B=15\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(3B=15\left(\frac{1}{1}-\frac{1}{100}\right)=15\left(\frac{100}{100}-\frac{1}{100}\right)=15.\frac{99}{100}\)

\(B=\frac{1}{3}.15-\frac{1}{3}.\frac{99}{100}=5-\frac{33}{100}=\frac{500}{100}-\frac{33}{100}=\frac{467}{100}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)

Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha

b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(2A=\frac{1}{3}-\frac{1}{51}\)

\(2A=\frac{16}{51}\)

\(A=\frac{16}{51}:2\)

\(A=\frac{8}{51}\)

5*(5-5/4+5/4-5/7+.......+5/100-5/103)

5*(5-5/103)

5*......... bạn tự tính nhé

câu b 1/3*5+1/5*7+............+1/49*51

1*(1/1-1/3+1/3-1/5+............+1/49-1/51)

1/1-1/51 tính ra rồi ra kết quả

tk nha