依题意得：αxβ=3m=4α+β=2m∴α²+β²=(2m)²-2x(3m+4)=4m²-6m+8又∵方程X²-2mx+3m+4=0有两个实数根∴Δ=4m²-12m-16≥0即m≥4或m≤-1∴当m=-1时,4m²-6m+8取得最小值4x(-1)²...