.5.7.11.13.37-10101/1212120+40404

=3.5.7.11.13.37-10101/1212120.1/10+40404 (vì 1/1212120=1/121212.1/10)

= 3.37.7.11.13.5-101010/121212.1/100+40404

=111.1001.5-5/6.1/100+40404

=151515.5-250/3

=595959-250/3

=1787876/3

1787876/3

A=3.5.7.11.13.37-10101/1212120+40404

A=(5.11).(3.7.13.37)-10101/10101.120+4.10101

A=55.10101-10101/10101.(120+4)

A=10101.(55-1)/10101.124

A=54/124

A=27/62

Ta có:

\(A=\frac{3.5.7.11.13.37-10101}{1212120+40404}\)

\(=\frac{\left(3.7.11.13.37\right).5-10101.1}{120.10101+4.10101}\)

\(=\frac{10101.\left(5-1\right)}{10101.\left(120+4\right)}\)

\(=\frac{4}{124}=\frac{1}{31}\)

Sai rồi:

A = 5.11.(3.7.13.37) - 10101/(10101.120 + 10101.4)

= (5.11.10101 - 10101)/(10101.120+10101.4)

= 10101(5.11-1)/10101(120+4)

= 27/62.

A= 5.11.(3.7.13.37)-10101/(10101.120+10101.4)

=(5.11.10101-10101)/(10101.120+10101.4)

=10101.(5.11-1)/10101(120+4)

=27/62

k mik nha mik k lại 100 cái lun

\(A=\frac{3.5.7.11.13.37-10101}{1212120+40404}\)

\(A=\frac{\left(3.7.13.17\right).\left(5.11\right)-10101}{120.10101+4.10101}\)

\(A=\frac{10101.55-10101.1}{120.10101+4.10101}\)

\(A=\frac{10101.\left(55-1\right)}{10101.\left(120+4\right)}\)

\(A=\frac{10101.54}{10101.124}\)

\(A=\frac{54}{124}\)

\(A=\frac{27}{62}\)

- chúc bn hok tốt

#Hạ 

A=3*5*7*11*13*37-10101/1212120+40404

A=10101*(55-1)/10101*(120+4)

A=55-1/120+4

A=54/124=27/62

A=\(\frac{\left(5.11\right)\left(3.7.13.37\right)-10101}{12.10101+4.10101}\)

A=\(\frac{55.10101-10101}{12.10101+4.10101}\)

A=\(\frac{54.10101}{16.10101}\)

A=\(\frac{54}{16}=3,375\)

A= [5.11.10101 -10101]/[10101.120+10101.4] = 10101.[5.11-1] / 101.[120+4] = 54/124=27/62

\(=\frac{27}{62}\)

tíck mik nha

A= [5.11.10101 -10101]/[10101.120+10101.4] = 10101.[5.11-1] / 101.[120+4] = 54/124=27/62

tính ra kết quả luôn nha các pạn ^ ^

\(A=\frac{3.5.7.11.13.37-10101}{1212120+40404}\)

\(=\frac{15015.37-273.37}{32760.37+1093.37}\)

\(=\frac{14742.37}{33853.37}\)

\(=\frac{14742}{33853}\)

\(A=\frac{3\cdot5\cdot7\cdot11\cdot13\cdot37-10101}{1212120+40404}\)

\(A=\frac{(3\cdot7\cdot11\cdot13\cdot37)\cdot5-10101}{10101\cdot120+10101\cdot4}\)

\(A=\frac{10101\cdot5-10101}{10101\cdot(120+4)}\)

\(A=\frac{10101\cdot(5-1)}{10101\cdot(120+4)}=\frac{4}{124}=\frac{1}{31}\)

\(\frac{27}{62}\)

\(\frac{27}{62}\)

a: \(\dfrac{31995-81}{42660-108}=\dfrac{31914}{42552}=\dfrac{3}{4}\)

b: \(\dfrac{3\cdot5\cdot7\cdot11\cdot13\cdot37-10101}{1212120+40404}=\dfrac{545454}{1252524}=\dfrac{27}{62}\)

a, $\frac{31995-81}{42660-108}$

= $\frac{395.81-81}{395.108-108}$

=$\frac{81.394}{394.108}$

= $\frac{81}{108}$          = $\frac{3}{4}$

b, $\frac{3.5.7.11.13.37-10101}{1212120+40404}$

= $\frac{3.5.7.11.13.37-3.7.13.37}{\mathrm{3.7.12.13.37.10}+\mathrm{3.4.7.13.37}}$

= $\frac{\left(\mathrm{3.7.13.37}\right)(5.11-1)}{\left(\mathrm{3.7.13.37}\right)(120+4)}$

= $\frac{5.11-1}{120+4}$

= $\frac{54}{124}$ = $\frac{27}{62}$