\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\)
\(=\dfrac{20}{35.31}+\dfrac{30}{35.41}+\dfrac{45}{50.41}+\dfrac{35}{50.57}\)
\(=5\left(\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\right)\)
\(=5\left(\frac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(B=\dfrac{7}{19.31}+\dfrac{5}{19.43}+\dfrac{3}{23.43}+\dfrac{11}{23.57}\)
\(=\dfrac{14}{34.31}+\dfrac{10}{38.43}+\dfrac{6}{46.43}+\dfrac{22}{46.57}\)
\(=2\left(\dfrac{7}{34.31}+\dfrac{5}{38.43}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{A}{B}=\frac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)
A=47.31+67.41+910.41+710.57A=47.31+67.41+910.41+710.57
=2035.31+3035.41+4550.41+3550.57=2035.31+3035.41+4550.41+3550.57
=5(435.31+635.41+950.41+750.57)=5(435.31+635.41+950.41+750.57)
=5(131−135+135−141+141−150+150−157)=5(131−135+135−141+141−150+150−157)
=5(131−157)=5(131−157)
B=719.31+519.43+323.43+1123.57B=719.31+519.43+323.43+1123.57
=1434.31+1038.43+646.43+2246.57=1434.31+1038.43+646.43+2246.57
=2(734.31+538.43+346.43+1146.57)=2(734.31+538.43+346.43+1146.57)
=2(131−134+134−143+143−146+146−157)=2(131−134+134−143+143−146+146−157)
=2(131−157)=2(131−157)
⇒AB=5131−1572131−157=52
\(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(=\frac{20\text{}}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)
\(=5\left(\frac{4\text{}}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}\right)\)
\(=5\left(\frac{1\text{}}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(=5\left(\frac{1\text{}}{31}-\frac{1}{57}\right)\)
\(B=\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\)
\(=\frac{14}{38.31}+\frac{10}{38.43}+\frac{6}{46.43}+\frac{22}{46.57}\)
\(=2\left(\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{46.43}+\frac{11}{46.57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{5\left(\frac{1}{31}-\frac{1}{57}\right)}{2\left(\frac{1}{31}-\frac{1}{57}\right)}=\frac{5}{2}\)
Vậy : \(\frac{A}{B}=\frac{5}{2}\)
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