因式分解:(1)x^2-y^2-z^2+2yz(2)-x^2+x+2(3)(x-1)(x-3)(x+1)(x+3)-20

如题所述

解:①x²-y²-z²+2yz=x²-(y²+z²-2yz)=x²-(y-z)²=(x+y-z)(x-y+z) ②-x²+x+2=-(x+1)(x-2)
③(x-1)(x-3)(x+1)(x+3)-20=(x²-9)(x²-1)-20=(x²)²-10x²-11=(x²+1)(x²-11)=(x²+1)(x+√11)(x-√11)追问

我们还没学(x²+1)(x+√11)(x-√11),可不可以最后一步就写(x²+1)(x²-11)

追答

(x²+1)(x²-11)为有理数范围内分解,(x²+1)(x+√11)(x-√11)为实数范围内分解

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