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\(A=3\cdot5+5\cdot7+7\cdot9+...+97\cdot99+99\cdot100=\)
\(7\cdot A=3\cdot5\cdot7+5\cdot7\cdot7+7\cdot9\cdot7+...+97\cdot99\cdot7+99\cdot101\cdot7=\)
\(7\cdot A=3\cdot5\cdot7+5\cdot7\cdot\left(9-2\right)+...+99\cdot101\cdot\left(103-96\right)=\)
\(7\cdot A=3\cdot5\cdot7+5\cdot7\cdot9+...+99\cdot101\cdot103-3\cdot5\cdot7-...-97\cdot99\cdot101=\)
\(7\cdot A=99\cdot101\cdot103=\)
\(A=\frac{99\cdot101\cdot103}{7}=...\)
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{4}{3\cdot5}+\frac{6}{5\cdot7}+\frac{8}{7\cdot9}+....+\frac{100}{99\cdot101}\)
\(=2\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{99\cdot100}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=2\cdot\frac{98}{101}\)
\(=\frac{196}{101}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{2}{99}-\frac{2}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{303}{303}-\frac{3}{303}=\frac{300}{303}\)
\(=\frac{100}{101}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{303}\)
Ta có:
\(C= 4+44+444+......+4444444444\)
\(C= 4.(10.1+9.10+8.100+7.1000+...+1.1000000000\)
\(C= 4.(100+90+800+7000+60000+500000+4000000+30000000+200000000+1000000000)\)
\(C=4.12345678900\)
\(C=4938271600\)
Tương tự.