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\(\frac{27^2.15^3.125^{17}.10^{10}.3^8.512+1024.25^{28}.12^6.15^7}{125.6^{17}+625^{12}.6^{13}.125^2.25^5.16}\)
=\(\frac{3^6.3^3.5^3.5^{34}.2^{10}.5^{10}.3^8.2^9+2^{10}.5^{56}.3^6.2^{12}.3^7.5^7}{5^{63}.6^{17}+5^{48}.6^{13}.5^6.2^4}\)=\(\frac{3^{17}.5^{47}.2^{19}+2^{22}.5^{63}.3^{13}}{5^{63}.6^{17}+5^{48}.6^{13}.5^6.2^4}=\frac{3^{13}.5^{47}.2^{19}.3^4+2^{19}.5^{47}.3^{13}.2^3.5^{16}}{5^{54}.6^{13}.5^9+5^{54}.6^{13}.2^4}\)
=\(\frac{3^{13}.5^{47}.2^{19}\left(3^4+2^3.5^{16}\right)}{5^{54}.6^{13}\left(5^9+2^4\right)}=\frac{6^{13}.2^6.5^{47}\left(3^4+2^3.5^{16}\right)}{5^7.5^{47}.6^{13}\left(5^9+2^4\right)}=\frac{2^6\left(3^4+2^3.5^{16}\right)}{5^7\left(5^9+2^4\right)}\)
Bạn hãy tự giải nốt
Bài làm
\(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(C=\frac{2.5^{21}.5-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{14}.7-19.7^{14}\right)}{7^{16}.7+3.7^{15}}\)
\(C=\frac{5^{21}.\left(2-9\right).5}{\left(5.5\right)^{10}}:\frac{5.[7^{15}.\left(3-19\right)].7}{7^{15}.\left(3+1\right).7}\)
\(C=\frac{5^{21}.\left(-7\right).5}{5^{10}.5^{10}}:\frac{5.7^{15}.\left(-16\right).7}{7^{15}.4.7}\)
\(C=\frac{5^{21}.\left(-35\right)}{5^{10}.5^{10}}:\frac{7^{15}.\left(-112\right).5}{7^{15}.28}\)
\(C=5.\left(-35\right):\frac{7^{15}.560}{7^{15}.28}\)
\(C=5.\left(-35\right):\frac{1.560}{1.28}\)
\(C=5.\left(-35\right):20\)
\(C=5.\left(-35\right).\frac{1}{20}\)
\(C=-\frac{175}{20}\)
\(C=-\frac{35}{4}\)
Vậy biểu thức trên \(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)bằng \(C=-\frac{35}{4}\)
# Chúc bạn học tốt #
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: \(E=\frac{50^{12}.16^{14}}{8^{25}.125^8}\)
\(E=\frac{\left(2.5^2\right)^{12}.\left(2^4\right)^{14}}{\left(2^3\right)^{25}.\left(5^3\right)^8}\)
\(E=\frac{2^{68}.5^{24}}{2^{75}.5^{24}}=\frac{1}{2^7}=\frac{1}{128}\)
a, \(-\frac{187}{70}\)
b,\(\frac{27}{70}\)
c,\(\frac{53}{14}\)
d,\(\frac{27}{4}\)
e,1
f,\(\frac{23}{4}\)
g,-1
i,6
k,315
l,\(\frac{9}{2}\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
\(\frac{25^{14}.5^{10}.625^3.125^7}{5^{14}.125^{10}.25^3.625^7}=\frac{\left(5^2\right)^{14}.5^{10}.\left(5^4\right)^3.\left(5^3\right)^7}{5^{14}.\left(5^3\right)^{10}.\left(5^2\right)^3.\left(5^4\right)^7}=\frac{5^{28}.5^{10}.5^{12}.5^{21}}{5^{14}.5^{30}.5^6.5^{28}}\)
\(=\frac{5^{71}}{5^{78}}=\frac{1}{5^7}=\frac{1}{78125}\)