Q=\(\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{1.99}}\)
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\(\frac{4+\frac{3}{5}+\frac{3}{7}+.....+\frac{3}{97}+\frac{3}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{99.1}}=???????\)
\(A=\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
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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Rút gọn:\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{99.1}}\)
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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\(\frac{1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+....+\frac{1}{49.51}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+......+\frac{1}{49.51}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{\frac{1+99}{1.99}+\frac{3+97}{3.97}+\frac{5+95}{5.95}+...+\frac{49+51}{49.51}}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{51}+...+\frac{1}{99}}\)
= 100
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tính GT của BT : \(A=\frac{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+....+\frac{1}{99.1}}\)
\(A=\frac{1+\frac{1}{3}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}}\)
\(=\frac{\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100}{2}=50\)
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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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Tính: A=\(\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)= ?
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+....+\frac{1}{49.51}}\)