Cho \(A= \frac{4}{{7.31}} + \frac{6}{{7.41}} + \frac{9}{{10.41}} + \frac{7}{{10.57}} \) và \(B = \frac{7}{{19.31}} + \frac{5}{{19.43}} + \frac{3}{{23.43}} + \frac{{11}}{{23.57}}\). Tính \(\frac{A}{B}\)?
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Lời giải:
Báo sai\(\begin{array}{l} \frac{A}{5} = \frac{4}{{35.31}} + \frac{6}{{35.41}} + \frac{9}{{50.41}} + \frac{7}{{50.57}} \Leftrightarrow \frac{A}{5} = \frac{1}{{31}} - \frac{1}{{35}} + \frac{1}{{35}} - \frac{1}{{41}} + \frac{1}{{41}} - \frac{1}{{50}} + \frac{1}{{50}} - \frac{1}{{57}} = \frac{1}{{31}} - \frac{1}{{57}}\\ \Leftrightarrow A = 5\left( {\frac{1}{{31}} - \frac{1}{{57}}} \right)\\ \frac{B}{2} = \frac{7}{{38.31}} + \frac{5}{{38.43}} + \frac{3}{{46.43}} + \frac{{11}}{{46.57}} \Leftrightarrow \frac{B}{2} = \frac{1}{{31}} - \frac{1}{{38}} + \frac{1}{{38}} - \frac{1}{{43}} + \frac{1}{{43}} - \frac{1}{{46}} + \frac{1}{{46}} - \frac{1}{{57}}\\ \Leftrightarrow \frac{B}{2} = \frac{1}{{31}} - \frac{1}{{57}} \Leftrightarrow B = 2\left( {\frac{1}{{31}} - \frac{1}{{57}}} \right)\\ \Rightarrow \frac{A}{B} = \frac{5}{2} \end{array}\)
