a) Ta có : 

990 = 2.3².5.11

2610=2.3².5.29

ƯCLN(990;2610) = 2.3².5=90

\(\Rightarrow\frac{990}{2610}=\frac{990:90}{2610:90}=\frac{11}{29}\)

Tương tự với câu b) 

\(a,\frac{990}{2610}=\frac{990:90}{2610:90}=\frac{11}{29}\)

\(b,\frac{374}{506}=\frac{374:2}{506:2}=\frac{187}{253}=\frac{187:11}{253:11}=\frac{17}{23}\)

\(c,\frac{3600-75}{8400-175}=\frac{3525}{8225}=\frac{3525:25}{8225:25}=\frac{141}{329}\)

\(d,\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{9^{12}.9^2.25^3.25^2.8^3.8^4}{\left(9.2\right)^{12}.\left(25.25\right)^3.\left(8.3\right)^3}=\frac{9^{12}.9^2.25^3.25^2.8^3.8^3.8}{9^{12}.2^{12}.25^3.25^2.25.8^3.3^3}=\frac{9^2.8}{2^{12}.25.3^3}\) đến đây bn tự làm nốt đi! 

~ mỏi ~

\(\frac{-72}{14}=\frac{-36}{7}\)

\(\frac{990}{2610}=\frac{11}{29}\)

\(\frac{374}{506}=\frac{17}{23}\)

\(\frac{3600-75}{8400-175}=\frac{3525}{8225}=\frac{3}{7}\)

hok tốt!!

dễ lắm mà

ừa, nhưng mik còn những câu sau khó hơn nhưng mỏi tay quá nên tí gửi tiếp!

\(\frac{990}{2610}=\frac{990:11}{2610:11}=\frac{11}{29}\)

\(\frac{374}{506}=\frac{374:22}{506:22}=\frac{17}{23}\)

\(\frac{3600-75}{8400-175}=\frac{3525}{8225}=\frac{3525:1175}{8225:1175}=\frac{3}{7}\)

\(\frac{9^{14}.255^5.8^7}{18^{12}.625^3.24^3}=\frac{3^{28}.5^5.3^5.17^5.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3^3}=\frac{3^{33}.5^5.17^5.2^{21}}{2^{21}.3^{27}.5^{12}}=\frac{3^6.17^5}{5^7}\)

a,999/2610=111/290

b,374/506=17/23

c,3600-75/8400-175=3/7

d,9^14.25^5.8^7/18^12.625^3.24^3=3/25

a, Ta có:999/2610=111/290

b, Ta có:374/506=17/2

c,Ta có:3600-75/8400-175=48.75-75/48.175-175=75.(48-1)/175.(48-1)=75/175=3/7

d, Ta có:(3^2)^14.(5^2)^5.(2^3)^7/(2.3^2)^12.(5^4)^3.(2^3.3)^3

=3^28.5^10.2^21/2^12.3^24.5^12.2^9.3^3

=3^28.5^10.2^21/2^21.3^27.5^12

=3/5^2=3/25

a, \(\dfrac{990}{2610}=\dfrac{11}{29}\)

b) \(\dfrac{3600-75}{8400-175}=\dfrac{3}{7}\)

c) \(\dfrac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^5\cdot24^3}=\dfrac{3}{25}\)

Cs thể ghi lời giải luôn đc hk bn.

\(A=\frac{3600-75}{8400-175}=\frac{3525}{8225}=\frac{3}{7}\)

\(B=\frac{9^{14}.25^7.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(5^2\right)^7.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}=\frac{3^{28}.5^{14}.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3^3}=\frac{3^{28}.5^{14}.2^{21}}{2^{21}.3^{27}.5^{12}}=3.5^2=3.25=75\)

\(C=\frac{1+3+5+...+19}{21+23+25+...+39}=\frac{\left(19+1\right).10:2}{\left(39+21\right).10:2}=\frac{100}{300}=\frac{1}{3}\)

\(D=\frac{71.52+53}{530.71-180}=\frac{3745}{37450}=\frac{1}{10}\).