Tính \(\frac{5^2.25^4}{125^3}\)
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0
11 tháng 8 2015
\(\frac{27^2.15^3.125^{17}.10^{10}.3^8.512+1024.25^{28}.12^6.15^7}{125.6^{17}+625^{12}.6^{13}.125^2.25^5.16}\)
=\(\frac{3^6.3^3.5^3.5^{34}.2^{10}.5^{10}.3^8.2^9+2^{10}.5^{56}.3^6.2^{12}.3^7.5^7}{5^{63}.6^{17}+5^{48}.6^{13}.5^6.2^4}\)=\(\frac{3^{17}.5^{47}.2^{19}+2^{22}.5^{63}.3^{13}}{5^{63}.6^{17}+5^{48}.6^{13}.5^6.2^4}=\frac{3^{13}.5^{47}.2^{19}.3^4+2^{19}.5^{47}.3^{13}.2^3.5^{16}}{5^{54}.6^{13}.5^9+5^{54}.6^{13}.2^4}\)
=\(\frac{3^{13}.5^{47}.2^{19}\left(3^4+2^3.5^{16}\right)}{5^{54}.6^{13}\left(5^9+2^4\right)}=\frac{6^{13}.2^6.5^{47}\left(3^4+2^3.5^{16}\right)}{5^7.5^{47}.6^{13}\left(5^9+2^4\right)}=\frac{2^6\left(3^4+2^3.5^{16}\right)}{5^7\left(5^9+2^4\right)}\)
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24 tháng 6 2017
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
24 tháng 6 2017
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
3 tháng 3 2018
a) -23 + 176 - (2176 - 23)
= -23 + 176 - 2176 + 23
= (-23 + 23) + (176 - 2176)
= 0 + (-2000)
= -2000
b) 125 . (-24) + 24 . 225
= (-125) . 24 + 24 . 225
= 24 . (-125 + 225)
= 24 . 100
= 2400
16 tháng 4 2020
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
27 tháng 4 2018
\(\frac{5-\frac{5}{7}-\frac{5}{49}}{4-\frac{4}{7}-\frac{4}{49}}+\frac{1,5+75\%-\frac{3}{8}}{0.625-\frac{5}{2}-125\%}\)
\(=\frac{5.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{49}\right)}{4\cdot\left(\frac{1}{4}-\frac{1}{7}-\frac{1}{49}\right)}+\frac{\frac{3}{2}+\frac{3}{4}-\frac{3}{8}}{\frac{5}{8}-\frac{5}{2}-\frac{5}{4}}\)
\(=\frac{5}{4}+\frac{3\cdot\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}\right)}{5\cdot\left(\frac{1}{8}-\frac{1}{2}-\frac{1}{4}\right)}\)
\(=\frac{5}{4}+\left(-\frac{3}{5}\right)\)
\(=\frac{13}{20}\)
Y
1
4 tháng 10 2021
\(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^{10}\cdot16}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^{10}\cdot2^4}=\dfrac{2^4\cdot5^3\cdot7}{2^{14}}=\dfrac{5^3\cdot7}{2^{10}}=\dfrac{875}{1024}\)
\(\frac{5^2.25^4}{125^3}=\frac{5^2.\left(5^2\right)^4}{\left(5^3\right)^3}=\frac{5^2.5^8}{5^9}=\frac{5^{10}}{5^9}=5.\)