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\(A=\frac{4^{15}.5^{15}.3^{12}.5^{12}}{5^{28}.2^{14}.3^{14}.2^{15}}=\frac{2.2^{29}.5^{27}.3^{12}}{5.5^{27}.2^{29}.3^{12}.3^2}=\frac{2}{45}\)
\(A=\frac{20^{15}.15^{12}}{25^{14}.6^{14}.8^5}=\frac{\left(2^2.5\right)^{15}.\left(3.5\right)^{12}}{\left(5^2\right)^{14}.\left(2.3\right)^{14}.\left(2^3\right)^5}=\frac{\left(2^2\right)^{15}.5^{15}.3^{12}.5^{12}}{5^{28}.2^{14}.3^{14}.2^{15}}\)
=\(\frac{2^{30}.3^{12}.5^{15}.5^{12}}{5^{28}.3^{14}.2^{14}.2^{15}}=\frac{2^{30}.3^{12}.5^{27}}{5^{28}.3^{14}.2^{29}}=\frac{2}{5.3^2}=\frac{2}{45}\)
\(\frac{25^{14}.5^{10}.625^3.125^7}{5^{14}.125^{10}.25^3.625^7}=\frac{\left(5^2\right)^{14}.5^{10}.\left(5^4\right)^3.\left(5^3\right)^7}{5^{14}.\left(5^3\right)^{10}.\left(5^2\right)^3.\left(5^4\right)^7}=\frac{5^{28}.5^{10}.5^{12}.5^{21}}{5^{14}.5^{30}.5^6.5^{28}}\)
\(=\frac{5^{71}}{5^{78}}=\frac{1}{5^7}=\frac{1}{78125}\)
\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{43}}\)
\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}\)
\(A=\frac{5^{30}.7^{48}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}=5.7^3.\left(1-7.2^2\right)=1715.\left(-27\right)=-46305\)
\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}\left(2^4\right)^2.7^{43}}=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}=\frac{7^{48}.5^{30}.2^8\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}\)
=\(7^5.5.\left(-27\right)=-2268945\)
Ta có
• A=1+34+38+312
=>34.A=34+38+312+316
<=>81.A-A=316-1
<=>A=(316-1)/80=538084
•B=1+32+34+36+38+310+312+314
=>32.B=32+34+36+38+310+312+314+316
<=>8.B=316-1
<=>B=(316-1)/8=53808400
Vậy Q=A/B=538084/53808400=1/100=0.01
f) \(\frac{25^2.20^4}{5^{10}.4^5}=\frac{\left(5^2\right)^2.\left(4.5\right)^4}{5^{10}.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{100}\)
g) \(\frac{16^{12}.8}{32^5.64^4}=\frac{\left(2^4\right)^{12}.2^3}{\left(2^5\right)^5.\left(2^6\right)^4}=\frac{2^{48}.2^3}{2^{25}.2^{24}}=\frac{2^{51}}{2^{49}}=2^2=4\)
h) \(\frac{2^{18}.9^4}{6^6.8^4}=\frac{2^{18}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^4}=\frac{2^{18}.3^8}{2^6.3^6.2^{12}}=\frac{2^{18}.3^8}{2^{18}.3^6}=3^2=9\)
c: \(C=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\dfrac{9^3}{4^3}:\dfrac{3^3}{16^3}}{2^7\cdot5^2+2^9}=\dfrac{1+1728}{3712}=\dfrac{1729}{3712}\)
\(D=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\dfrac{3^5-3^4}{3^6+3^5}=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}=\dfrac{2}{3\cdot4}=\dfrac{2}{12}=\dfrac{1}{6}\)
\(E=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}=\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}=5\cdot\dfrac{-2}{3}=\dfrac{-10}{3}\)
ta có: \(E=\frac{50^{12}.16^{14}}{8^{25}.125^8}\)
\(E=\frac{\left(2.5^2\right)^{12}.\left(2^4\right)^{14}}{\left(2^3\right)^{25}.\left(5^3\right)^8}\)
\(E=\frac{2^{68}.5^{24}}{2^{75}.5^{24}}=\frac{1}{2^7}=\frac{1}{128}\)
ta có: \(E=\frac{50^{12}.16^{14}}{8^{25}.125^8}\)
\(E=\frac{\left(2.5^2\right)^{12}.\left(2^4\right)^{14}}{\left(2^3\right)^{25}.\left(5^3\right)^8}\)
\(E=\frac{2^{68}.5^{24}}{2^{75}.5^{24}}=\frac{1}{2^7}=\frac{1}{128}\)