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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
\(\frac{3600-75}{8400-175}=\frac{75.48-75}{175.48-175}=\frac{75.\left(48-1\right)}{175.\left(48-1\right)}\)
\(=\frac{75}{175}=\frac{25.3}{25.7}=\frac{3}{7}\)
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!
\(\frac{44116-14}{10290-35}=\frac{44102}{10255}\)
\(\frac{3600-75}{8400-175}=\frac{75\left(48-1\right)}{35\left(240-5\right)}=\frac{15\cdot47}{7\cdot235}=\frac{3\cdot5\cdot47}{7\cdot5\cdot47}=\frac{3}{7}\)
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) \(\dfrac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^5\cdot24^3}=\dfrac{3}{25}\)
\(A=\frac{3600-75}{8400-175}=\frac{3525}{8225}=\frac{3}{7}\)
\(B=\frac{9^{14}.25^7.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(5^2\right)^7.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}=\frac{3^{28}.5^{14}.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3^3}=\frac{3^{28}.5^{14}.2^{21}}{2^{21}.3^{27}.5^{12}}=3.5^2=3.25=75\)
\(C=\frac{1+3+5+...+19}{21+23+25+...+39}=\frac{\left(19+1\right).10:2}{\left(39+21\right).10:2}=\frac{100}{300}=\frac{1}{3}\)
\(D=\frac{71.52+53}{530.71-180}=\frac{3745}{37450}=\frac{1}{10}\).