2/1.3+2/3.5+2/5.7+...+2/101.103=1/1-1/3+1/3-1/5+1/5-1/7+...+1/101-1/103
=1-1/103=102/103
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Các câu hỏi tương tự
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot12}\)
Tính nhanh:
A = \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{2001\cdot2003}+\frac{2}{2003\cdot2005}\)
\(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\cdot\left(x-\frac{1}{2}\right)=\frac{3}{4}\)
bài 7:tính bằng cách thuận tiện nhất
a, 1/3+2/3...+3\(\frac{1}{3}\) + 3\(\frac{2}{3}\) =.......
b,9+9/2+9/4+9/8......+9/128+9/256=......
c,\(\frac{5}{3}\) + \(\frac{5}{3\cdot5}\) + \(\frac{5}{5\cdot7}\) + ......... +\(\frac{5}{101\cdot103}\) =.........
\(S=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
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 :
\(\frac{2}{5}< A< \frac{8}{9}\)
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
rút gọn
\(\frac{2^5\cdot7\cdot2^5}{2^5\cdot5^2-25\cdot3}\)
\(a,A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(b\)
\(ab\cdot10-ab=2ab\)