\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\right)\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{101}{303}-\frac{3}{303}=\frac{98}{303}\)
Đặt A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Leftrightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.100}\)
\(=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{7}+\frac{1}{9}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
Ta có:
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{303}\)
Ta có:
2/3x5 + 2/5x7 + 2/7x9 + ... + 2/99x101
=1/3 - 1/5 + 1/5 - 1/7 + ... + 1/97 - 1/99
=1/3 - 1/99
=32/99
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+.....+\left(\frac{-1}{99}+\frac{1}{99}\right)-\frac{1}{101}\)
\(\frac{1}{3}-\frac{1}{101}=\frac{101}{303}-\frac{3}{303}=\frac{98}{303}\)
\(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+......+\frac{2}{99x101}\)
\(=2\left(\frac{1}{3x5}+\frac{1}{5x7}+....+\frac{1}{99x101}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=2x\frac{98}{303}\)
\(=\frac{196}{303}\)
1/2×(2/3×5+2/5×7+2/7×9+.....+2/99×101)
=> 1/3×5+1/5×7+.....+.1/99×101
1/3-1/5+1/5-1/7 +.....+ 1/99-1/101
1/3-1/101 =98/303
ra kết quả là \(\frac{98}{303}\)
kết bạn nha
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{303}\)
