分解-a(a-x)(x-b) ab(a-x)(b-x)
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(a-b)(x-y)-2(a-b)(x+y)(a-b)(x-y-2x-2y)(a-b)(-x-3y)
x^2-3ax+2a^2-ab-b^2=0x^2-3ax+(2a+b)(a-b)=0[x-(2a+b)][x-(a-b)]=0x1=2a+bx2=a-
x^2(a-b)+4(b-a)=x^2(a-b)-4(a-b)=(x^2-4)(a-b)=(x+2)(x-2)(a-b)
a(y-x)-2b(x-y)=a(y-x)+2b(y-x)=(y-x)(a+2b)
=(x+a)(x+4a)(x+2a)(x+3a)+a^4=(x^2+5ax+4a^2)(x^2+5ax+6a^2)+a^4=(x^2+5ax)^2+10a^2(x^2+5ax)+25a^4=(x^2+
a^2(a-b)+(b-a)=(a²-1)·(a-b)=(a+1)·(a-1)·(a-b)(x-y)^2-2(y-x)+1=(x-y)^2+2(x-y)+1=[(x-y)+1]²=
法1(2a+b)(2a-3b)-3a(2a+b)=2a(2a-3b)+b(2a-3b)-6a²-3ab=4a²-6ab+2ab-3b²-6a²-3ab=-2a&
x(a-x)(a-y)-y(x-a)(y-a)=(x-y)(x-a)(y-a)-ab(a-b)^2+a(b-a)^2=(a-b)^2(-ab+a)=a(a-b)^2(1-b)a^3b-ab=ab(a^
p(x-y)-q(y-x)=p(x-y)+q(x-y)=(p+q)(x-y)(a+b)(a-b)-(b+a)=(a+b)(a-b-1)10a(x-y)^2-5b(y-x)=10a(x-y)^2+5b(
(x+2)(x+6)=x^2+8x+12a=8(x+1)(x+16)=x^2+17x+16b=163^n+2-4×3^n+1+10×3^n=3^2*3^n-4*3*3^n+10*3^n=3^n(9-1
你的分母中有两个因式为x和(x-1)^2
=[2(a-b)]^2-(x-y)^2=(2a-2b+x-y)(2a-2b-x+y)
x(x-y)(a-b)-y(y-x)(b-a),=x(x-y)(a-b)-y(x-y)(a-b),=(x-y)(a-b)(x-y),=(x-y)2(a-b).
/>1.a(x+y)-b(x+y)=(x+y)(a-b)2.m(a-b)-n(b-a)=m(a-b)+n(a-b)=(a-b)(m+n)
(b-a)^2=(a-b)^2嘛,所以:=(a-b)^2(x-1)
分解为ax+ay+bx+by-ax+ay-bx+by=ay+by+ay+by=2ay+2by=2y(a+b)
解题思路:(1)利用平方差公式分解因式;(2)把(b-a)转化为-(a-b)然后再提取公因式即可.解题过程:
(x+a)(x+b)这个直接乘出来=x^2+bx+ax+ab=x^2+(a+b)x+abx^2+(a+b)x+ab=x^2+ax+bx+ab=x(x+a)+b(x+a)=(x+b)(x+a)
解由题设可得x²-3x+a=(x-5)(x+b)令x=5,可得10+a=0a=-10令x=0,可得-10=a=-5b∴b=2∴a=-10,b=2