\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\Rightarrow\dfrac{A}{5}=\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\)

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

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

\(\dfrac{B}{2}=\dfrac{7}{38.31}+\dfrac{5}{38.41}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\Rightarrow\dfrac{B}{2}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\)

\(\Rightarrow B=\dfrac{52}{31.57}\)

\(\dfrac{A}{B}=\dfrac{130}{31.51}:\dfrac{52}{31.57}=\dfrac{5}{2}\)

Ta có : `A:5=4/35.31+6/35.41++9/50.41+7/50.57`

`A/5 = 1/31 - 1/35 +1/35-1/41+1/41-1/50+1/50-1/57`

`A/5 =1/31-1/57` `(1)`

Lại có : `B:2=7/38.31 +5/38.43+3/46.43+11/46.57`

`B/2 =1/31-1/38+1/38-1/43+1/43-1/46+1/46-1/57`

`B/2 =1/31-1/57` `(2)`

Từ `(1)` và `(2)` `=>A/5 =B/2`

                         `=>A/B=5/2`

a) Ta có: \(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)

\(=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)

\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)

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

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

Ta có: \(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)

\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)

\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)

\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)

\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)

Suy ra: \(\dfrac{A}{B}=\dfrac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)

giúp mình với mai thi rùi

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

`Answer:`

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

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

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

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

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

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

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

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

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

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

A = \(\frac{4}{7}.31+\frac{6}{7}.41+\frac{9}{10}.41+\frac{7}{10}.57\)

= \(\left[\left(\frac{4}{7}+\frac{6}{7}\right).\left(31+41\right)\right]+\left[\left(\frac{9}{10}+\frac{7}{10}\right).\left(41+57\right)\right]\)

= \(\frac{10}{7}.72+\frac{8}{5}.98\)

= \(\left(\frac{10}{7}+\frac{8}{5}\right).\left(72+98\right)\)

= \(\left(\frac{50}{35}+\frac{56}{35}\right).170\)

= \(\frac{106}{35}.170\)

= \(\frac{3604}{7}\)

B = \(\frac{7}{19}.31+\frac{5}{19}.43+\frac{3}{23}.43+\frac{11}{23}.57\)

= \(\left[\left(\frac{7}{19}+\frac{5}{19}\right).\left(31+43\right)\right]+\left[\left(\frac{3}{23}+\frac{11}{23}\right).\left(43+57\right)\right]\)

= \(\frac{12}{19}.74+\frac{14}{23}.100\)

= \(\left(\frac{12}{19}+\frac{14}{23}\right).\left(100+74\right)\)

= \(\left(\frac{276}{437}+\frac{266}{437}\right).174\)

= \(\frac{542}{437}.174\)

= \(\frac{79674}{437}\)

A : B = \(\frac{3604}{7}:\frac{19674}{437}=\frac{3604.437}{7.19674}=\frac{1802.2.437}{7.9837.2.}=\frac{1802.437}{7.9837}\)

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:

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

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

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

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

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