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

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

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

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

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

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

=> \(Q=\frac{300}{2}=150\)

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 \(A=50\)

Đặt \(\frac{A}{B}=\frac{4+\frac{3}{5}+\frac{3}{7}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{4+\frac{3}{5}+\frac{3}{7}+...+\frac{3}{93}+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+\frac{1}{7.93}+...+\frac{1}{93.7}+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{4+3.\frac{1}{5}+3.\frac{1}{7}+...+3.\frac{1}{93}+3.\frac{1}{95}+3.\frac{1}{97}+3.\frac{1}{99}}{1.\frac{1}{99}+\frac{1}{3}.\frac{1}{97}+\frac{1}{5}.\frac{1}{95}+\frac{1}{7}.\frac{1}{93}+...+\frac{1}{93}.\frac{1}{7}+\frac{1}{95}.\frac{1}{5}+\frac{1}{97}.\frac{1}{3}+\frac{1}{99}.1}\)

\(\Leftrightarrow\frac{A}{B}=\frac{4+3+3+...+3+3+3+3}{1.\frac{1}{99}+\frac{1}{3}.\frac{1}{97}+...+\frac{1}{93}.\frac{1}{7}+\frac{1}{95}.\frac{1}{5}.\frac{1}{3}.1}\)

P/s:Tới đây bạn giải tiếp nha! Mình cũng không chắc cho lắm! Khi nào mình biết mình sẽ giải tiếp cho bạn! Nên đừng dis

Câu này mình chưa được học mà! Bạn cứ giải tiếp đi xem nào!

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 => \(A=\frac{C}{A}=50\)

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 \(A=50\)

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