1)(a^2-3a+2)x^2+(2a^2-4a+1)xy+(a^2-a)y^2
=(a-1)(a-2)x^2+(2a^2-4a+1)x+a(a-1)y^2
=[(a-1)x+ay][(a-2)x+(a-1)y]
2)3x^2y^2 -17mnxy+10m^2n^2
=(3xy-2mn)(xy-5mn)
3)x^2+4y^2+9z^2-4xy-12yz+6xz
=(x-2y)^2+6z(x-2y)+9z^2
=(x-2y+3z)^2
4)a2-b2-ac+c2/4
=(a-c/2)^2-b^2
=(a+b-c/2)(a-b-c/2)
5)-2x?5n*y?n+4x?3n*y?n+2-2x?n*y?n+4
=-2x^ny^n(x^4n-2x^2ny^2+y^4)
=-2x^ny^n(x^2n-y^2)^2
1.十字相乘法
2.公式法
3.分組分解法
4.主元法:
5.構(gòu)造法
6.換元法
需要原理及過程,我不僅想知道過程更想知道怎么做
需要原理及過程,我不僅想知道過程更想知道怎么做
1)(a²-3a+2)x²+(2a²-4a+1)xy+(a²-a)y²
2)3x²y²-17mnxy+10m²n²
3)x²+4y²+9z²-4xy-12yz+6xz
4)(a²-3a+2)x²+(2a²-4a+1)xy+(a²-a)y²
5)-2xˆ5n-1yˆn+4xˆ3n-1yˆn+2-2xˆn-1yˆn+4
有一題打錯了,補上:
4)a²-b²-ac+四分之一c²
1)(a²-3a+2)x²+(2a²-4a+1)xy+(a²-a)y²
2)3x²y²-17mnxy+10m²n²
3)x²+4y²+9z²-4xy-12yz+6xz
4)(a²-3a+2)x²+(2a²-4a+1)xy+(a²-a)y²
5)-2xˆ5n-1yˆn+4xˆ3n-1yˆn+2-2xˆn-1yˆn+4
有一題打錯了,補上:
4)a²-b²-ac+四分之一c²
數(shù)學(xué)人氣:493 ℃時間:2020-05-24 01:44:52
優(yōu)質(zhì)解答
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