(y+z)(x﹣y)(x+z) D. (y+z)(x+y)(x﹣z) 相关知识点: 试题来源: 解析 [解答]解:x2y﹣y2z+z2x﹣x2z+y2x+z2y﹣2xyz =(y﹣z)x2+(z2+y2﹣2yz)x+z2y﹣y2z =(y﹣z)x2+(y﹣z)2x﹣yz(y﹣z) =(y﹣z)[x2+(y﹣z)x﹣yz] =(y﹣z)(x+y)(x﹣z). 故选:...
解析 [解答] 解: x2y- y2z+z2x - x2z+y2x+z2y- 2xyz =( y- z) x2+( z2+y2 - 2yz) x+z2y - y2z =( y- z) x2+( y- z) 2x- yz( y- z) 2 =( y- z) [ x2+( y- z) x- yz] =( y- z)( x+y)( x- z). 故选: A....
解答:x2y-y2z+z2x-x2z+y2x+z2y-2xyz=(y-z)x2+(z2+y2-2yz)x+z2y-y2z=(y-z)x2+(y-z)2x-yz(y-z)=(y-z)[x2+(y-z)x-yz]=(y-z)(x+y)(x-z).故选A.点评:本题考查了用分组分解法进行因式分解,难点是将原式重新整理成关于x的二次三项式,改变其结构,寻找分解的突破口. ...
原式=(y-z)x2+(y2-2yz+z2)x-(zy2-z2y) =(y-z)[x2+(y-z).x-yz] =(y-z)(x+y)(x-z). 结果一 题目 【例2】分解多项式 :x^2y-y^2z+z^2x-x^2z+y^2x+z^2y-2xyz. 答案 【解答】原式 =(y-z)x^2+(y^2-2yz+z^2)x-(zy^2-z^2y)=(y-z)[x^2+(y-z)x-yz]=(...
(matrix)=( y-z )(x^2)+((( y-z ))^2)x+yz( z-y ) =( y-z )[ (x^2)+( y-z )x-yz ] =( y-z )( x+y )( x-z ) (matrix)结果一 题目 用主元法分解因式:⑴x2y−y2z+z2x−x2z+y2x+z2y−2xyzA.(y+z)(x+y)(x−z)B.(y−z)(x+y)(x−z)...
用主元法分解因式:(1)x2y−y2z+z2x−x2z+y2x+z2y−2xyz.(2)x2y−y2z+z2x−x2z+y2x+z2y−2xyz. 答案 (1)(y夜zs(x+y子(x−z表.hal尽y−z)(x+容)s数−zs.相关推荐 1用主元法分解因式:(1)x2y−y2z+z2x−x2z+y2x+z2y−2xyz.(2)x2y−y2z+...
百度试题 结果1 题目【题目】因式分解:x2y-y2z+z2x-x2z+y2x+z2y-2xyz. A.(xy+yz-xz)2 B.(y-x)(x-z)(z+y) C.(y-z)(x+y)(x-z) D.(y-z)(x-y)(x-z) 相关知识点: 试题来源: 解析 【解析】C 反馈 收藏
已知x,y,z都是正整数,代数式x2y−y2z+z2x−x2z+y2x+z2y−2xyz的值是质数,则zx+y的值是( ). A.0B.1C.2D.4
相关知识点: 试题来源: 解析 B(2)原式=(y-z)x2+(z2+y2-2y)x+z2y-y2z-|||-=(y-z)z2+(y-z)x+yz(z-y)-|||-=(y-z)[x2+(y-z)x-y-|||-=(y-z)(x+y)(x-z)· 反馈 收藏
x2y−y2z+z2x−x2z+y2x+z2y−2xyz=(y−z)(x+y)(x−z), 由于结果为质数,且x+y⩾2,故x−z=y−z=1或x−z=y−z=−1, 若x−z=y−z=1,则x+y=2z+2=2(z+1)为合数,舍去, 若x−z=y−z=−1,则x+y=2z−2=2(z−1), 故z=2,x=y=1,zx+...