\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=\frac{1}{5}-\frac{1}{20}=\frac{3}{20}\)
A=\(A=\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot25\cdot35}{1\cdot5\cdot7+2\cdot10\cdot19+4\cdot20\cdot28+7\cdot35\cdot49}\)
* là dấu nhân
\(S=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH :
\(\frac{2}{5}< A< \frac{8}{9}\)
tinh tong A+\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3+4}+...\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(a,A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(b\)
\(ab\cdot10-ab=2ab\)
so sánh\(\frac{1\cdot3\cdot5+2\cdot6+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)với \(\frac{303}{708}\)
\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+\(\frac{1}{4\cdot5}\)+\(\frac{1}{5\cdot6}\)+\(\frac{1}{6\cdot7}\)+\(\frac{1}{7\cdot8}\)+\(\frac{1}{8\cdot9}\)+ \(\frac{1}{9\cdot10}\)
Tính hộ tôi [Lưu ý : dấu chấm bằng dấu nhân]
so sánh A và B biết
A=\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
B=\(\frac{111111}{666665}\)
