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a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}\)
\(=\frac{1}{10}\)
b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)
\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)
\(=\frac{1}{10}-\frac{1}{1000}\)
\(=\frac{99}{1000}\)
c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)
\(=4.\left(1-\frac{1}{90}\right)\)
\(=4.\frac{89}{90}\)
\(=\frac{178}{45}\)
_Chúc bạn học tốt_
\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)
\(\Rightarrow5A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)
\(\Rightarrow5A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\right)\)
\(\Rightarrow5A=1-\frac{1}{8}\)
\(\Rightarrow A=\left(1-\frac{1}{8}\right).\frac{1}{5}=\frac{7}{40}\)
\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{5}{7.8}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=5\left(1-\frac{1}{8}\right)\)
\(A=5.\frac{7}{8}\)
\(A=\frac{38}{8}\)
Bài làm:
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{9.10}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)=2.\frac{4}{10}=\frac{4}{5}\)
\(A=\frac{1}{2\times3}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{98\times99}\)
\(=\frac{1}{6}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}+...+\frac{1}{98\times99}\)
\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{99}=\frac{161}{396}>\frac{160}{400}=\frac{2}{5}\)
A= 5.(1/5.6+1/6.7+...+1/10.11)
A=5.(1/5-1/6+1/6-1/7+.....+1/10-1/11)
A=5.(1/5-1/11)
A=5.6/55=6/11
A=6/11<1
A<1