Đề : \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{80}+\dfrac{1}{120}\)

tính Bình thường

​\(A=\dfrac{240+30+10+3+2}{240}=\dfrac{285}{240}=\dfrac{57}{48}=\dfrac{19}{16}\)

các PA cho bạn chọn

PA.1 ;

\(A=1+\left(\dfrac{10}{8.10}+\dfrac{1}{80}\right)+\left(\dfrac{5}{24.5}+\dfrac{1}{120}\right)=1+\left(\dfrac{10+1}{80}+\dfrac{5+1}{120}\right)\)

\(A=1+\left(\dfrac{11}{80}+\dfrac{1}{20}\right)=1+\dfrac{11+4}{80}=1+\dfrac{15}{80}=1+\dfrac{3}{16}=\dfrac{19}{16}\)

PA2.

\(A=1+\dfrac{1}{8}\left(1+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{15}\right)=1+\dfrac{1}{8}\left(\dfrac{4}{3}+\dfrac{3+2}{30}\right)=1+\dfrac{1}{8}\left(\dfrac{4}{3}+\dfrac{1}{6}\right)\)

\(A=1+\dfrac{1}{8}\left(\dfrac{8+1}{6}\right)=1+\dfrac{1}{8}.\dfrac{3}{2}=\dfrac{16+3}{16}=\dfrac{19}{16}\)