若x^2-3x+1=0,求代数式4x^2-[3x^2+2(x^2-x)-4x-5]的值

问题描述:

若x^2-3x+1=0,求代数式4x^2-[3x^2+2(x^2-x)-4x-5]的值
明天要交了

x^2-3x+1=0,(x-1)(2x-1)=0,x1=1,x2=1/24x^2-[3x^2+2(x^2-x)-4x-5]=4x^2-[3x^2+2x^2-2x-4x-5]=4x^2-[5x^2-6x-5]=4x^2-5x^2+6x+5=-x^2+6x+5=-(x^2-6x-5)①=-(x^2-6x-5)=-[1-6-5]=10②=-(x^2-6x-5)=-[1/4-3-5]=8-1/4=7...