\(\text{Ta có:}\)

\(B=\frac{1.3.5.....49}{26.27.....50}=\frac{1.2.3.....50}{26.27...50}.\frac{1}{2.4.....50}=\frac{1.2.3....25}{1.2.3....25.2^{25}}=\frac{1}{2^{25}}\)

\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)

\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)

\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)

\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)

3.3 va 2,8 nhe chuc ban thanh cong nho ket ban voi to nhe

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\(\frac{1.3.5....49}{27.28.29...50}=\frac{1.3.5....\left(27.29...49\right)}{\left(27.29...49\right).\left(28.30...50\right)}=\frac{1.3.5....25}{28.30....50}\)=\(\frac{13}{4^32^6.8.16.32}=\frac{13}{2^6.2^6.2^3.2^4.2^5}=\frac{13}{2^{24}}\)

\(\frac{1.3.5....49}{26.27...50}\)

= \(\frac{1.3.5...49.2.4.6...50}{26.27...50.2.4.6...50}\)

=\(\frac{1.2.3.4.5.....49.50}{1.2.3.4.5.....49.50.2^{25}}\)

= \(\frac{1}{2^{25}}\)