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三次方的因式分解,要详细的过程
x³-3x²+4=
ab(c²-d²)+cd(a²-b²)=
x²-4mx+8mn-4n⁴=
x⁴+64=
x³-11x²+31x-21=
x³-4xy²-2x²y+8y³=
人气:338 ℃ 时间:2019-11-06 23:53:36
解答
1、原式=(x³-x²)-(4x²-4)=x²(x-1)-4(x+1)(x-1)=(x-1)(x²-4x-4)
2、原式=abc²-abd²+a²cd-b²cd=(abc²+a²cd)-(abd²+b²cd)=ac(bc+ad)-bd(ad+bc)=(ad+bc)(ac-bd)
3、原式=(x²-4mx+4m²)-(4m²-8mn+4n²)=(x-2m)²-(2m-2n)²
=(x-2m+2m-2n)(x-2m-2m+2n)=(x-2n)(x-4m+2n)--------题目有错,该是:-4n²
4、原式=(x²+8)²-16x²=(x²+4x+8)(x²-4x+8)
5、原式=(x³-x²)-(10x²-10x)+(21x-21)=x²(x-1)-10x(x-1)+21(x-1)=(x-1)(x²-10x+21)=(x-1)(x-3)(x-7)
6、原式=(x³-2x²y)-(4xy²-8y³)=x²(x-2y)-4y²(x-2y)=(x-2y)(x²-4y²)=(x-2y)(x+y)(x-y)
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