Question

（x'2-y'2-1）'2-4y'2分式因解

Answer 1

直接运用平方差公式 原式子=(x^2-y^2-1-2y)(x^2-y^2-1+2y)=[x^2-(y^2+2y+1)][x^2-(y^2-2u+1)]=

[x^2-(y+1)^2][x^2-(y-1)^2]=(x-y-1)(x+y+1)(x-y+1)(x+y-1)