Question

若（x^m÷x^2n)^3÷x^(m-n)与-1/3x^2为同类项,且2m+5n=6.求4m^2-25n^2的平方根

Answer 1

(x^m÷x^2n)^3÷x^(m-n)

=[x^(m-2n)]^3÷x^(m-n)

=x^[3(m-2n)-(m-n)]

=x^(2m-5n)

同类项则x的次数相等

所以2m-5n=2

又2m+5n=6

所以4m^2-25n^2=(2m+5n)(2m-5n)=2*6=12