`Answer:`

\(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)

\(=\frac{5.4}{5.\left(7.31\right)}+\frac{5.6}{5.\left(7.41\right)}+\frac{5.9}{5.\left(10.41\right)}+\frac{5.17}{5.\left(10.57\right)}\)

\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)

\(=5.\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}\right)\)

\(=5.\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)

\(=5.\left(\frac{1}{31}-\frac{1}{57}\right)\)

\(=5.\frac{26}{1767}\)

\(=\frac{130}{1767}\)

\(B=\frac{4}{31.7}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)

\(\frac{B}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)

\(\frac{B}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)

\(B=5\left(\frac{1}{31}-\frac{1}{57}\right)=5\cdot\frac{26}{1767}=\frac{130}{1767}\)

=4/217+6/287+9/410+7/570

=130/1767

A= 

7.31

4

​

 + 

7.41

6

​

 + 

10.41

9

​

 + 

10.57

7

​

 ⇒ 

5

A

​

 = 

35.31

4

​

 + 

35.41

6

​

 + 

50.41

9

​

 + 

50.57

7

​

\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57} 

5

A

​

 = 

31

1

​

 − 

35

1

​

 + 

35

1

​

 − 

41

1

​

 + 

41

1

​

 − 

50

1

​

 + 

50

1

​

 − 

57

1

​

\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57} 

5

A

​

 = 

31

1

​

 − 

57

1

​

 = 

31.57

26

​

 ⇒A= 

31.57

130

​

T viết nhầm

Hơhow

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=\(\frac{39}{434}\) 

Ta có:

\(\frac{B}{5}=\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}+\frac{13}{57.70}\)

\(=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}+\frac{1}{57}-\frac{1}{70}\)

\(=\frac{1}{31}-\frac{1}{70}\)

\(\Rightarrow B=5\left(\frac{1}{31}-\frac{1}{70}\right)\)

\(=\frac{39}{434}\)

\(\frac{A}{5}=\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}=\frac{35-31}{35.31}+\frac{41-35}{35.41}+\frac{50-41}{50.41}+\frac{57-50}{50.57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)=> A = 5. \(\left(\frac{1}{31}-\frac{1}{57}\right)\)

\(\frac{B}{2}=\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{38-31}{31.38}+\frac{43-38}{38.43}+\frac{46-43}{43.46}+\frac{57-46}{46.57}\)

\(\frac{B}{2}=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)=> B = 2.\(\left(\frac{1}{31}-\frac{1}{57}\right)\)

A/B = 5/2