Question 308978
In this problem, remember that {{{(x^(y))/(x^(z))=x^(y-z)}}}. In other words, when you divide like bases with exponents, you subtract the exponents. 



Also, keep in mind that negative exponents basically tell you to 'flip' the fraction of the base.



{{{(2x^2)/(7x^7)=(2/7)((x^2)/(x^7))=(2/7)x^(2-7)=(2/7)x^(-5)=(2/7)(1/x^5)=2/(7x^5)}}}


So in short, {{{(2x^2)/(7x^7)=2/(7x^5)}}} where {{{x<>0}}}