Question

已知m^2+9n^2+4m-6n+5=0,求m,n的值

当m,n取第(1)小题的值时，求（2m-3n)^2-(2m+3n)(2m-3n)+(2m+3n)^2的值。

Answer 1

m^2+9n^2+4m-6n+5=0,
m^2+4m+4+9n^2-6n+1=0,
(m+2)²+(3n-1)²=0,
(m+2)²>=0,(3n-1)²>=0,
所以(m+2)²=0,m=-2;(3n-1)²=0,n=1/3;

（2m-3n)^2-(2m+3n)(2m-3n)+(2m+3n)^2
=[（2m-3n)-(2m+3n)]²+(2m+3n)(2m-3n)
=[2m-3n-2m-3n)]²+(2m)²-(3n)²
=[-6n)]²+(2m)²-(3n)²
=[-6/3)]²+(-4)²-(1)²
=4+16-1
=19
或
（2m-3n)^2-(2m+3n)(2m-3n)+(2m+3n)^2
=（-4-1)^2-(-4+1)(-4-1)+(-4+1)^2
=25-15+9
=19