Lời giải:

$\frac{125^3.7^5-175^5}{2016^{2017}}=\frac{7^5.(125^3-25^5)}{2016^{2017}}=\frac{7^5.(25^3.5^3-25^5)}{2016^{20117}}=\frac{7^5.25^3.(5^3-25^2)}{2016^{2017}}=\frac{7^5.5^6.(5^3-5^4)}{2016^{2017}}=\frac{(-4).5^3.5^6.7^5}{2016^{2017}}=\frac{(-4).5^9.7^5}{2016^{2017}}=\frac{(-4).5^9.7^5}{7^{2017}.288^{2017}}=\frac{(-4).5^9}{7^{2012}.288^{2017}}=\frac{-2^2.5^9}{7^{2012}.2^{2017}.144^{2017}}=\frac{-5^9}{7^{2012}.2^{2015}.144^{2017}}=\frac{-5^9}{7^{2012}.2^{2015}.16^{2017}.9^{2017}}=\frac{-5^9}{7^{2012}.9^{2017}.2^{10083}}=0$(Vì tử quá nhỏ mẫu rất lớn nên phân thức bằng $0$)

Lời giải:

1253.75−175520162017=75.(1253−255)20162017=75.(253.53−255)201620117=75.253.(53−252)20162017=75.56.(53−54)20162017=(−4).53.56.7520162017=(−4).59.7520162017=(−4).59.7572017.2882017=(−4).5972012.2882017=−22.5972012.22017.1442017=−5972012.22015.1442017=−5972012.22015.162017.92017=−5972012.92017.210083=01253.75−175520162017=75.(1253−255)20162017=75.(253.53−255)201620117=75.253.(53−252)20162017=75.56.(53−54)20162017=(−4).53.56.7520162017=(−4).59.7520162017=(−4).59.7572017.2882017=(−4).5972012.2882017=−22.5972012.22017.1442017=−5972012.22015.1442017=−5972012.22015.162017.92017=−5972012.92017.210083=0(Vì tử quá nhỏ mẫu rất lớn nên phân thức bằng 00)