关于x的一元二次方程x^2-x+p-1=0有两个实数根x1x2,若｛2+x1(1-x1)}{2+x2{1-x2}=9,求P的值

由韦达定理得：x1+x2=1x1x2=p-1故9=[2+x1(1-x1)][2+x2(1-x2)]=4+2(x1-x1^2+x2-x2^2)+x1x2(1-x1)(1-x2)=4+2[(x1+x2)-(x1+x2)^2+2x1x2]+x1x2[1-(x1+x2)+x1x2]=4+2[1-1+2(p-1)]+(p-1)[1-1+p-1]=4+4(p-1)+(p-1)^2=(p-1+2...