1) Tính:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+........+\frac{2}{97.99}\)
\(B=\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.......\frac{1}{97.99}\)
\(D=\left(\frac{\frac{12}{1}-\frac{12}{5}-\frac{12}{487}}{\frac{4}{1}-\frac{4}{5}-\frac{4}{487}}\right):\left(\frac{\frac{5}{1}+\frac{5}{17}+\frac{5}{37}}{\frac{6}{1}+\frac{6}{17}+\frac{6}{37}}\right).\frac{505}{1818}\)
2) So sánh:
a)\(\frac{2008}{2009}+\frac{2009}{2010}\) và \(\frac{2008+2009}{2009+2010}\)
b) \(\frac{12}{23}\) và \(\frac{1212}{2323}\)
Giúp mình nha. Mai mình kiểm tra 45'.
B = \(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5.1}{1.1.1}=10\)
C = \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}\left(\frac{33}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
1) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
