\(A=\dfrac{4^2.25^2+32.125}{\dfrac{2^3.5^2+49^{25}.125^{10}.2^8+5^{30}.\left(-7\right)^{49}.4^4}{16^2.5^{29}.\left(-7\right)^{48}}}\)
\(A=\dfrac{\left(2^2\right)^2.\left(5^2\right)^2+2^5.5^3}{\dfrac{2^3.5^2+\left(7^2\right)^{25}.\left(5^3\right)^{10}.2^8+5^{30}.\left(-7\right)^{49}.\left(2^2\right)^4}{\left(2^4\right)^2.5^{29}.\left(-7\right)^{48}}}\)
\(A=\dfrac{2^4.5^4+2^5.5^3}{\dfrac{2^3.5^2+7^{50}.5^{30}.2^8+5^{30}.\left(-7\right)^{49}.2^8}{2^8.5^{29}.\left(-7\right)^{48}}}\)
\(A=\dfrac{2^4.5^2.\left(5^2+2\right)}{\dfrac{2^3.5^2+7.\left(7^{49}.5^{30}.2^8-7^{49}.5^{30}.2^8\right)}{2^8.5^{29}.7^{48}}}\)
\(A=\dfrac{\dfrac{2^4.5^2.\left(25+2\right)}{2^3.5^2}}{2^8.5^{29}.7^{48}}\)
\(A=\dfrac{2^4.5^2.27}{2^3.5^2}\div2^8.5^{29}.7^{48}\)
\(A=\dfrac{2^4.5^2.3^3}{2^3.5^2.2^8.5^{29}.7^{48}}\)
\(A=\dfrac{27}{2^7.5^{29}.7^{48}}\)
Vậy....
