若方程x∧2－2（m+1）x+3m∧2－4mn+4n∧2+2=0有实根,则有：Δ≥0即：4(m+1)²-4(3m²-4mn+4n²+2)≥0(m+1)²-(3m²-4mn+4n²+2)≥0m²+2m+1-3m²-4mn-4n²-2≥0-m²+2m-1-m...