Ta có : \(\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{89.90}\)
\(=\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+.......+\frac{1}{89}-\frac{1}{90}\)
\(=\frac{1}{3}-\frac{1}{90}=\frac{30}{90}-\frac{1}{90}=\frac{29}{90}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{89.90}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{89}-\frac{1}{90}\)
\(=\frac{1}{3}-\frac{1}{90}\)
\(=\frac{29}{90}\)
Đặt \(A=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{89.90}\)
\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{89}-\frac{1}{90}\)
\(A=\frac{1}{3}-\frac{1}{90}\)
\(A=\frac{29}{90}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{89.90}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{89}-\frac{1}{90}\)
= \(\frac{1}{3}-\frac{1}{90}\)
= \(\frac{29}{90}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{89.90}\)
\(=\frac{4-3}{3\cdot4}+\frac{5-4}{4\cdot5}+...+\frac{90-89}{89\cdot90}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{89}-\frac{1}{90}\)
\(=\frac{1}{3}-\frac{1}{90}=\frac{29}{90}\)
1/3.4 + 1/4.5 + ... + 1/89.90
= 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/89 - 1/90
= 1/3 - 1/90
= 29/90
Đs:
Ps: Sao câu này có nhiều ctv trả lời thế nhỉ?