如果x₁、x₂是关于x的一元二次方程x²+ax+a=2的两个实根

根据韦达定理,可得：x1+x2=-ax1x2=a-2(x1-2x2)(x2-2x1)=x1x2-2x1²-2x2²+4x1x2=5x1x2-2(x1²+x2²)=5x1x2-2[(x1+x2)²-2x1x2]=5x1x2-2(x1+x2)²+4x1x2=9x1x2-2(x1+x2)²=9(a-2)-2a²=-2a²+9a-18=-2[a²-(9/2)a]-18=-2[a²-(9/2)a+(9/4)²]-18+2×(9/4)²=-2[a-(9/4)]²-63/8方程有两个实数根,则△≥0△=a²-4(a-2)=a²-4a+8=(a-2)²+4无论a取何实数值,△﹥0(x1-2x2)(x2-2x1)在当a=9/4时,有最大值-63/8