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Ta có:10000-25/1200-3
=400.25-25/400.3-3
=25.(400-1)/3.(400-1)
=25.399/3.399
=25/3
\(\frac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}=\frac{\left(3^2\right)^{14}\cdot\left(5^2\right)^5\cdot\left(2^3\right)^7}{\left(3^2\cdot2\right)^{12}\cdot\left(5^4\right)^3\cdot\left(3\cdot2^3\right)^3}\)
\(=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{24}\cdot2^{12}\cdot5^{12}\cdot3^3\cdot2^9}=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{25}\cdot5^{12}\cdot2^{21}}=\frac{3^3}{5^2}=\frac{27}{25}\)
Đặt phân số trên là A
\(A=\frac{25^{28}+25^{24}+...+25^4+25^0}{\left(25^{28}+25^{24}+...+25^4+25^0\right)+\left(25^{30}+25^{26}+...+25^6+25^2\right)}\)
\(\frac{1}{A}=\frac{\left(25^{28}+25^{24}+...+25^4+25^0\right)+\left(25^{30}+25^{26}+...+25^6+25^2\right)}{25^{28}+25^{24}+...+25^4+25^0}\)
\(\frac{1}{A}=1+\frac{25^{30}+25^{26}+...+25^6+25^2}{25^{28}+25^{24}+...+25^4+25^0}\)
Đặt \(B=\frac{25^{30}+25^{26}+...+25^6+25^2}{25^{28}+25^{24}+...+25^4+25^0}\)
\(\frac{B}{25^2}=\frac{25^{30}+25^{26}+...+25^6+25^2}{25^{30}+25^{26}+...+25^6+25^2}=1\Rightarrow B=25^2\)
=> \(\frac{1}{A}=1+B=1+25^2\Rightarrow A=\frac{1}{1+25^2}\)
a) \(\frac{\left(-63\right)}{81}=\frac{\left(-63\right):9}{81:9}=\frac{-7}{9}\)
b) \(\frac{-25}{-75}=\frac{\left(-25\right):25}{\left(-75\right):25}=\frac{-1}{-3}=\frac{1}{3}\)
\(a.\frac{-63}{81}=\frac{-7}{9}\)
\(b.\frac{-25}{-75}=\frac{1}{3}\)
\(\frac{10000+25}{1200+3}=\frac{10025}{1203}=\frac{25}{3}\)