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Ta có : \(\frac{\frac{1}{39}-\frac{1}{6}-\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\left(\frac{1}{39}-\frac{1}{6}-\frac{1}{51}\right)\times5304}{\left(\frac{1}{8}-\frac{1}{52}+\frac{1}{68}\right)\times5304}=\frac{136-884-104}{663-102+78}=\frac{-852}{639}=-\frac{4}{3}\)
Xét : 1/8 - 1/52 + 1/68
= 3/4 . 1/6 - 3/4 . 1/39 + 3/4 . 1/68
= 3/4 . (1/6-1/39+1/51)
=> E = 1/(3/4) = 4/3
Tk mk nha
\(\text{E}=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
\(\text{E}=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{3}{4}.\frac{1}{6}-\frac{3}{4}.\frac{1}{39}+\frac{3}{4}.\frac{1}{68}}\)
\(\text{E}=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{3}{4}\left(\frac{1}{6}-\frac{1}{39}+\frac{1}{51}\right)}\)
\(\text{E}=\frac{1}{\left(\frac{3}{4}\right)}=\frac{4}{3}\)
Đề hơi sai đó vì mình dùng máy tính bấm ra dc:\(\frac{-5112}{19747}\)
\(M=\left(1-\frac{1}{2}\right)-\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{6}\right)-....-\left(1-\frac{1}{200}\right)\)
\(M=-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-.....-\frac{1}{200}\right)=-\frac{1}{2}\left(1-\frac{1}{2}+...-\frac{1}{100}\right)\)
Xét:
\(S=1-\frac{1}{2}+....-\frac{1}{100}.S=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+....+\frac{1}{100}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\Rightarrow M=-\frac{1}{2}\left(\frac{1}{51}+....+\frac{1}{100}\right)\)
N:M=-2
\(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+......-\frac{1}{49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right);B=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-.....-\frac{1}{49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\Rightarrow\frac{A}{B}=\frac{17}{13}\)
Có: \(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Ttự, ta đc: \(B=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Vậy \(\frac{A}{B}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
#Walker
\(B=\frac{\frac{1}{39}-\frac{1}{6}-\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}:\frac{31}{6}\)
\(=\frac{\frac{1}{3}\left(\frac{1}{13}-\frac{1}{2}-\frac{1}{17}\right)}{\frac{1}{4}\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}.\frac{6}{31}\)
\(=\frac{\frac{-1}{3}\left(\frac{-1}{13}+\frac{1}{2}+\frac{1}{17}\right)}{\frac{1}{4}\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}.\frac{6}{31}\)
\(=\frac{-1}{3}:\frac{1}{4}.\frac{6}{31}\)
\(=\frac{-1}{3}.4.\frac{6}{31}\)
Tiếp theo dễ r tự làm tiếp :)