`Answer:`

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

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

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

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

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

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

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

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

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

=130/1767

Đáp án là \(\frac{64}{403}\)bạn nhé !

A= 5 x (4/31.7.5 + 6/7.5.41 + 9/41.10.5 + 7/ 10.5.57 + 1958/57.403.5)

   = 5 x (4/31.35 + 6/35.41 + 9/41.50 + 7/50.57 + 1958/57.2015)

   =5 x (1/31 - 1/35 + 1/35 - 1/41 + 1/41 - 1/50 + 1/50 - 1/57 + 1/57 - 1/2015)

   = 5 x (1/31 - 1/2015)

   = 5 x (65/2015 - 1/2015)

   = 5 x 64/2015

   = 320/2015 = 64/403

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

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

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

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

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

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

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