使关于x的方程(3x^2-2m^2+mn)^2+(3m^2-mn+2n^2-12x)^2+4=4x-x^2至少有一个实数解的实数对(m,n)的个数是

(3x^2-2m^2+mn)^2+(3m^2-mn+2n^2-12x)^2+4=4x-x^2所以(3x^2-2m^2+mn)^2+(3m^2-mn+2n^2-12x)^2+4+x^2-4x=0所以(3x^2-2m^2+mn)^2+(3m^2-mn+2n^2-12x)^2+(x-2)^2=0所以三个平方都等于0所以3x^2-2m^2+mn=03m^2-mn+2n^2-1...