Ta có: \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}\)
\(B=\frac{10}{10}-\frac{1}{10}\)
\(B=\frac{9}{10}\)
Vậy: \(B=\frac{9}{10}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}\)
\(B=\frac{9}{10}\)
Vì \(\frac{9}{10}< 1\)nên B < 1
Vậy B < 1
Ta có
\(B=\frac{1}{2}+\frac{1}{6} +...+\frac{1}{90}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}\)
Vì 0<\(\frac{1}{10}\)<1
=>1-\(\frac{1}{10}\)<1
=>B<1
Ta có: B = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{72}+\frac{1}{90}\)
B= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
B= \(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}\)+...+\(\frac{9-8}{8.9}+\frac{10-9}{9.10}\)
B= \(\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{9}{8.9}-\frac{8}{8.9}+\frac{10}{9.10}-\frac{9}{9.10}\)
B=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+....+\frac{1}{8}\)\(-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
B= \(1-\frac{1}{10}\)
B= \(\frac{9}{10}\)
Do 9<10 => B< 1
Duyệt đi, chúc bạn học giỏi !
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\) và 1
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\Rightarrow B< 1\)
