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Ta có: A=47.31 +67.41 +910.41 +710.57
=2035.31 +3035.41 +4550.41 +3550.57
=5(431.35 +635.41 +941.50 +750.57 )
=5(131 −135 +135 −141 +141 −150 +150 −157 )
=5(131 −157 )
B=719.31 +519.43 +323.43 +1123.57
=1438.31 +1038.43 +646.43 +2246.57
=2(731.38 +538.43 +343.46 +1146.57 )
=2(131 −138 +138 −143 +143 −146 +146 −157 )
=2(131 −157 )
⇒AB =5(131 −157 )2(131 −157 ) =52
Ta có: \(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)
\(=5\left(\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(B=\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\)
\(=\frac{14}{38.31}+\frac{10}{38.43}+\frac{6}{46.43}+\frac{22}{46.57}\)
\(=2\left(\frac{7}{31.38}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{5\left(\frac{1}{31}-\frac{1}{57}\right)}{2\left(\frac{1}{31}-\frac{1}{57}\right)}=\frac{5}{2}\)
