计算Δ=04(m+1)²-4(3m²+4mn+4n²+2)=02m²+4mn+4n²-2m+1=0(m-1)²+(m+2n)²=0m=1且m=-2n所以m=1，n=-1/2
△=04（m+1)^2-4(3m^2+4mn+4n^2+2)=0-8m^2+8m-16mn-16n^2-4=0-4m^2-16mn-16n^2-4m^2+8m-4=04(m+2n)^2+4(m-1)^2=0m=1,n=-1/2
