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 $0$)
