F=4/7.3+6/7.4+9/10.41+7/10.57+13/57.14
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`Answer:`
\(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(=\frac{5.4}{5.\left(7.31\right)}+\frac{5.6}{5.\left(7.41\right)}+\frac{5.9}{5.\left(10.41\right)}+\frac{5.17}{5.\left(10.57\right)}\)
\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)
\(=5.\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}\right)\)
\(=5.\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(=5.\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(=5.\frac{26}{1767}\)
\(=\frac{130}{1767}\)
Đây là toán lớp 6.
=>1/5B= 4/7.5.31 +6/7.5.41+9/5.10.41+7/5.10.57+13/57.5.14
=>1/5B=4/31.35+6/35.41+....+13/57.70
=>1/5B=1/31-1/35+1/35-1/41+...+1/57-1/70
=>1/5B=1/31-1/70
=>1/5B=39/2170
=>B=39/2170:1/5
=>B=39/424
Ta có:
\(\frac{B}{5}=\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}+\frac{13}{57.70}\)
\(=\frac{35-31}{35.31}+\frac{41-35}{35.41}+\frac{50-41}{50.41}+\frac{57-50}{50.57}+\frac{70-57}{57.70}\)
\(=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}+\frac{1}{57}-\frac{1}{70}\)
\(=\frac{1}{31}-\frac{1}{70}\)
\(\rightarrow B=5\cdot\left(\frac{1}{31}-\frac{1}{70}\right)\)
\(=5\cdot\frac{39}{2170}\)
\(=\frac{39}{434}\)
Vậy B=\(\frac{39}{434}\)
\(B=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}+\frac{65}{57.70}\)
\(B=5.\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}+\frac{13}{57.70}\right)\)
\(B=5.\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}+\frac{1}{57}-\frac{1}{70}\right)\)
\(B=5.\left(\frac{1}{31}-\frac{1}{70}\right)\)
\(B=5.\frac{39}{2170}\)
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