Question

分解因式 （a+b)^2-4(a+b-1) x^2+8x(x+1)+16(x+1)^2 (x^2-x)^2+2分之1(x

Answer 1

(a+b)^2-4(a+b-1) =(a+b)^2-4(a+b)+4=(a+b-2)^2 x^2+8x(x+1)+16(x+1)^2 =[x+4(x+1)]^2=(5x+4)^2(x^2-x)^2+2分之1(x^2-x）+16分之1=[(x^2-x)+1/4 ]^2=[(x-1/2)^2]^2=(x-1/2)^4