Ta có: \(\frac{9^9.225^5.8^7}{18^{12}.625^3.24^3}\)=\(\frac{\left(3^2\right)^9.\left(25.3^2\right)^5.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(25^2\right)^3.\left(3.8\right)^3}\)

=\(\frac{3^{18}.25^5.3^{10}.2^{21}}{2^{12}.3^{24}.25^6.3^3.2^9}\)

=\(\frac{3^{28}.25^5.2^{21}}{3^{27}.25^6.2^{21}}\)=\(\frac{3}{25.}\)

\(\frac{9^{14}.225^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(3^2.5^2\right)^5.\left(2^3\right)^7}{\left(3^2.2\right)^{12}.\left(5^4\right)^3.\left(3.2^3\right)^3}=\frac{3^{28}.3^{10}.5^{10}.2^{21}}{3^{24}.2^{12}.5^{12}.3^3.2^9}=\frac{3^{38}.5^{10}.2^{21}}{3^{27}.2^{21}.5^{12}}=\frac{3^{11}}{5^2}\)

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

\(\frac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}=\frac{\left(3^2\right)^{14}\cdot\left(5^2\right)^5\cdot\left(2^3\right)^7}{\left(3^2\cdot2\right)^{12}\cdot\left(5^4\right)^3\cdot\left(3\cdot2^3\right)^3}\)

\(=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{24}\cdot2^{12}\cdot5^{12}\cdot3^3\cdot2^9}=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{25}\cdot5^{12}\cdot2^{21}}=\frac{3^3}{5^2}=\frac{27}{25}\)

\(\frac{9^{14}}{18^{12}}.\frac{25^5}{625^3}.\frac{8^7}{24^3}\)

\(=\frac{9^{14}}{\left(9.2\right)^{12}}.\frac{25^5}{25^6}.\frac{8^7}{\left(8.3\right)^3}\)

\(=\frac{9^{14}}{9^{12}.2^{12}}.\frac{1}{25}.\frac{8^7}{8^3.3^3}\)

\(=\frac{9^2}{2^{12}}.\frac{1}{25}.\frac{8^4}{3^3}\)

\(=\frac{81}{4096}.\frac{1}{25}.\frac{4096}{27}\)

\(=\frac{81}{4096}.\frac{4096}{27}.\frac{1}{24}=3.\frac{1}{24}=\frac{3}{24}\)

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trả lời 

A=3/24

suy ra A=1/8

20112-(304+2012)+(2013+304)

=20112-304-2012+2013+304

=20112+(-2012+2013)+(-304+304)

=20112+1+0=20113

\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.25^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(25^2\right)^3.\left(2^3\right)^3.3^3}=\)\(\frac{3^{28}.25^5.2^{21}}{2^{12}.2^9.3^{24}.3^3.25^6}=\frac{3^{28}.25^5.2^{21}}{2^{21}.3^{27}.25^6}\)\(=\frac{3}{25}\)

a = \(\frac{-8^3.6^4}{18^{12}}\)

   = \(\frac{-2^9.2^4.3^4}{2^{12}.3^{24}}\)

   = \(\frac{-2^{13}.3^4}{2^{12}.3^{24}}\)

   = \(\frac{-2}{3^{20}}\)

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