Ta có:
\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)
\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)
\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)
b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)
\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)
\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)
\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)
Bn Tấn sai rùi
phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)
Ta có: \(\frac{1}{20.21}=\frac{21-20}{20.21}=\frac{21}{20.21}-\frac{20}{20.21}=\frac{1}{20}-\frac{1}{21}\)
Suy ra: \(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)
\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)
\(=\frac{1}{20}-\frac{1}{61}=\frac{41}{1220}\)
Phép tính sau làm tương tự.
Ket qua phan a la \(\frac{41}{1220}\), phan b la\(\frac{44}{245}\)
\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)
=\(\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)
=\(\frac{1}{20}-\frac{1}{61}=\frac{41}{1220}\)
\(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)
=\(4.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\right)\)
=\(4.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\right)\)
=\(4.\left(\frac{1}{5}-\frac{1}{49}\right)\)
=\(4.\frac{44}{245}\)
=\(\frac{176}{245}\)
