分解因式(1)8a3b2-12ab3c(2)-3ma3+6ma2-12ma(3)2(x-y)2-x(x-y)(4)3ax2-6axy+3ay2(5)p2-5p-
题型:解答题难度:一般来源:不详
分解因式 (1)8a3b2-12ab3c (2)-3ma3+6ma2-12ma (3)2(x-y)2-x(x-y) (4)3ax2-6axy+3ay2 (5)p2-5p-36 (6)x5-x3 (7)(x-1)(x-2)-6 (8)a2-2ab+b2-c2 |
答案
(1)8a3b2-12ab3c=4ab2(2a2-3bc); (2)-3ma3+6ma2-12ma=-3ma(a2-2a+4)=-3ma(a-2)2; (3)2(x-y)2-x(x-y)=(x-y)(2x-2y-x)=(x-y)(x-2y); (4)3ax2-6axy+3ay2=3a(x2-2xy+y2)=3a(x-y)2; (5)p2-5p-36=(p-9)(p+4); (6)x5-x3=x3(x2-1)=x3(x+1)(x-1); (7)(x-1)(x-2)-6=x2-3x+2-6=(x-4)(x+1); (8)a2-2ab+b2-c2=(a-b)2-c2=(a-b+c)(a-b-c). |
举一反三
分解因式 (1)a2-1+b2-2ab (2)(x2+1)2-4x2 (3)(x+y)2+6(x+y)+9 (4)x4-5x2+4 |
已知a=b且a≠0. ①将等号两边各乘上a,得a2=ab; ②在等式两边各减去b2,得a2-b2=ab-b2; ③将上式分解因式,得(a-b)(a+b)=(a-b)b; ④在等式两边同时除以公因式(a-b),得a+b=b; ⑤已知a=b,所以2a=a; ⑥因为a≠0,所以可以推出2=1.上述推论是错误的,问题出在( ) |
下列各题中,分解因式正确的是( )A.b(a-4)-c(4-a)=(a-4)(b-c) | B.x2(x-2)2+2x(x-2)2=(x-2)2(x2+2x) | C.(a-b)(a-c)+(b-a)(b-c)=(a-b)(a+b-2c) | D.5a(x-y)+10b(y-x)=5(x-y)(a-2b) |
|
最新试题
热门考点