(x'2-y'2-1)'2-4y'2分式因解

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(x''2-y''2-1)''2-4y''2分式因解(x''2-y''2-1)''2-4y''2分式因解(x''2-y''2-1)''2-4y''2分式因解直接运用平方差公式原式子=(x^2-y^2-1-2y)(x^2-y^

(x'2-y'2-1)'2-4y'2分式因解
(x'2-y'2-1)'2-4y'2分式因解

(x'2-y'2-1)'2-4y'2分式因解
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)