dấu mũ ngược đời v

\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)

=\(\frac{3.2^3}{5^2}\)=24/25

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

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

\(\frac{3600-75}{8400-175}=\frac{75.48-75}{175.48-175}=\frac{75.\left(48-1\right)}{175.\left(48-1\right)}\)

\(=\frac{75}{175}=\frac{25.3}{25.7}=\frac{3}{7}\)

Trừ r dùng máy tính

Trả lời :

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

Tk mình nha

Chỉ còn cách này nữa thôi

\(\frac{3600-75}{8400-175}=\frac{75.48-75}{175.48-175}\)

\(=\frac{75.\left(48-1\right)}{175.\left(48-1\right)}\)

\(=\frac{75}{175}=\frac{3}{7}\)

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

\(=\frac{3}{7}\)

Chúc bạn học tốt nha!

còn cách nào nhanh hơn không???

\(\frac{44116-14}{10290-35}=\frac{44102}{10255}\)

\(\frac{3600-75}{8400-175}=\frac{75\left(48-1\right)}{35\left(240-5\right)}=\frac{15\cdot47}{7\cdot235}=\frac{3\cdot5\cdot47}{7\cdot5\cdot47}=\frac{3}{7}\)

g, \(\frac{44116-14}{10290-35}=\frac{44102}{10255}\)

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

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

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

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

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

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

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

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