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