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

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

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

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)\(\frac{\text{3600−75}}{\text{8400−105}}=\frac{3.1200-3.25}{7.1200-7.25}=\frac{3.\left(1200-25\right)}{7.\left(1200-25\right)}=\frac{3}{7}\)

b) \(\frac{9^{14}.25^5.8^2}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(5^2\right)^5.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{27}.5^{12}}=\frac{3}{5^2}=\frac{3}{25}\)

c) \(C=\frac{1+3+5+...+19}{21+23+25+...+39}=\frac{\left(1+19\right).19:2}{\left(21+39\right).19:2}=\frac{190}{570}=\frac{1}{3}\)

c)

C = 1+3+5+...+1921+23+25+...+39

Số số hạng của tử là : ( 19 -1 ):2 + 1= 10 số

Số số hạng của mẫu là : ( 39 - 21 ) : 2 +1 = 10 số

Tổng các số hạng của tử là : ( 1 + 19 ) x 10 : 2 = 100

Tổng các số hạng của mẫu là : ( 21 + 39 ) x 10 : 2 = 300

Vậy C = 100/300 = 1 / 3

Mik làm mỗi ý C) thui . Sorry nha ! Nhớ tik

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

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

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

\(\dfrac{9^{14}.25^6.8^7}{18^{12}.625^3.24^3}=\dfrac{\left(3^2\right)^{14}.\left(5^2\right)^6.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}\)

\(=\dfrac{3^{28}.5^{12}.2^{21}}{2^{21}.3^{27}.5^{12}}=\dfrac{3}{1}=3\)

\(A=\dfrac{9^{14}.25^6.8^7}{18^{12}.625^3.24^3}=\dfrac{\left(3^2\right)^{14}.\left(5^2\right)^6.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(3.2^3\right)^3}\)

=\(\dfrac{3^{28}.5^{12}.2^{21}}{2^{12}.3^{24}.5^{12}.3^3.2^9}\)=\(\dfrac{3^{28}.5^{12}.2^{21}}{2^{21}.3^{27}.5^{12}}=3\)