50 nha                          

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

Tính ở tử số: 

\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}=50.2.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)

\(=50.\left(\frac{1}{1.99}+\frac{1}{3.97}+.....+\frac{1}{49.51}+\frac{1}{51.49}+...+\frac{1}{99.1}\right)\)

Gọi tử số là C: mẫu số là B => \(A=\frac{C}{A}=50\)

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A= (2018+1) x 2018 : 2 = 2037171 

B= ko biết làm

Q=\(\frac{3+1+\frac{3}{5}+...+\frac{3}{99}}{\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)}\)

Q=\(\frac{\frac{3}{1}+\frac{3}{3}+\frac{3}{5}+...+\frac{3}{99}}{\frac{2}{1.99}+\frac{2}{3.97}+...+\frac{2}{49.51}}\)

Q=\(50.\frac{3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)}{50\left(\frac{2}{1.99}+\frac{2}{3.97}+...+\frac{2}{49.51}\right)}\)

Q=\(50.3.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}\)

Q=\(150.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{99+1}{1.99}+\frac{97+3}{3.97}+...+\frac{51+49}{49.51}}\)

Q=150\(.\frac{\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}\)

Q=\(150.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}\)

Q=150.1

Q=150

      \(Q=\frac{4+\frac{3}{5}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}}\)

=> \(Q=\frac{100\left(\frac{3}{1}+\frac{3}{3}+\frac{3}{5}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}\right)}{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}\right)}\)

=> \(Q=\frac{100.3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{\frac{1+99}{1.99}+\frac{3+97}{3.97}+\frac{5+95}{5.95}+...+\frac{95+5}{95.5}+\frac{97+3}{97.3}+\frac{99+1}{99.1}}\)

=> \(Q=\frac{300\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{95}+\frac{1}{5}\right)+\left(\frac{1}{97}+\frac{1}{3}\right)+\left(\frac{1}{99}+\frac{1}{1}\right)}\)

=> \(Q=\frac{300\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}\)

=> \(Q=\frac{300}{2}=150\)