Đáp án:
`A=50`
Giải thích các bước giải:
$\rm@sbpro2009$
$A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}\ \!\,\;\!\!\!\,\!+\ \!\,\ \!\!\!.\!.\!.\,\;\!\!+\ \dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\times99}+\dfrac{1}{3\times97}+\dfrac{1}{5\times 95}\ \!\!\,+\ \!.\!.\!.+\ \dfrac{1}{97\times 3}+\dfrac{1}{99\times1}}$
$A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+\left(\dfrac{1}{5}+\dfrac{1}{95}\right)+\ \! \!.\!.\!.+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{\left(\dfrac{1}{1\times99}+\dfrac{1}{99\times1}\right)+\left(\dfrac{1}{3\times97}+\dfrac{1}{97\times3}\right)+\ \!\!.\!.\!.+\left(\dfrac{1}{49\times51}+\dfrac{1}{51\times49}\right)}$
$A=\dfrac{\left(\dfrac{99}{99}+\dfrac{1}{99}\right)+\left(\dfrac{97}{3\times97}+\dfrac{3}{97\times3}\right)+\left(\dfrac{95}{5\times95}+\dfrac{5}{95\times5}\right)+\ \! \!.\!.\!.+\left(\dfrac{51}{49\times51}+\dfrac{49}{51\times49}\right)}{\dfrac{2}{1\times99}+\dfrac{2}{3\times97}+\dfrac{2}{5\times95}+\ \!\!.\!.\!.+\ \dfrac{2}{49\times51}}$
$A=\dfrac{\dfrac{100}{1\times99}+\dfrac{100}{3\times97}+\dfrac{100}{5\times95}\ +\,.\!.\!.+\ \dfrac{100}{49\times51}}{\dfrac{2}{1\times99}+\dfrac{2}{3\times97}+\dfrac{2}{5\times95}+\ \!\!.\!.\!.+\ \dfrac{2}{49\times51}}$
$A=\dfrac{100\times\left(\dfrac{1}{1\times99}+\dfrac{1}{3\times97}+\dfrac{1}{5\times95}+\ \!\!.\!.\!.+\ \dfrac{1}{49\times51}\right)}{2\times\left(\dfrac{1}{1\times99}+\dfrac{1}{3\times97}+\dfrac{1}{5\times95}+\ \!\!.\!.\!.+\ \dfrac{1}{49\times51}\right)}$
$A=\dfrac{100}{2}$
$A=50$
Vậy `A=50`.
