解:原式=(x2+xy)-(xz+yz)=x(x+y)-z(x+y)=(x+y)(x-z) .故答案为: (x+y)(x-z). 本题主要考查的是分组分解法分解因式,解决本题的关键是正确进行分组,然后提公因式进行因式分解. 本题主要考查的是分组分解法分解因式,解决本题的关键是正确进行分组,然后提公因式进行因式分解.难度不大,是一道...
解题思路：当被分解的式子是四项时，应考虑运用分组分解法进行分解．本题前两项、后两项都有公因式，且分解后还能继续分解，故使前两项一组，后两项一组．x2-xy+xz-yz，=（x2-xy）+（xz-yz），=x（x-y）+z（x-y），=（x-y）（x+z）．点评：本题考点： 因式分解-分组分解法．考点...
x2-xy+xz-yz,=(x2-xy)+(xz-yz),=x(x-y)+z(x-y),=(x-y)(x+z). APP内打开 为你推荐 查看更多 分解因式:x2-xy+xz-yz=_. 解答:解:x2-xy+xz-yz,=(x2-xy)+(xz-yz),=x(x-y)+z(x-y),=(x-y)(x+z). 23011 因式分解1.x^2+xy-yz-xz x^2+xy-yz-xz=x(x+y)-z(x...
解答:解:x2-xy+xz-yz, =(x2-xy)+(xz-yz), =x(x-y)+z(x-y), =(x-y)(x+z). 点评:本题考查用分组分解法进行因式分解.难点是采用两两分组还是三一分组.本题前两项、后两项都有公因式,且分解后还能继续分解,故使前两项一组,后两项一组. ...
解答:解:x2-xy+xz-yz, =(x2-xy)+(xz-yz), =x(x-y)+z(x-y), =(x-y)(x+z). 点评:本题考查用分组分解法进行因式分解.难点是采用两两分组还是三一分组.本题前两项、后两项都有公因式,且分解后还能继续分解,故使前两项一组,后两项一组. ...
前两项组合提一个X，后两项组合提一个Z 变为X（X-Y）+Z(X-Y)在提个（X-Y）变为（X-Y)(X+Z)就这样
x2-xz+xy-yz =(x2+xy)+(-xz-yz)=x(x+y)-z(x+y)=(x+y)(x-z)后面一题，基本因式分解 (x-7)(x+5)所以x+5
Description When I'm attaching TransformControls to an object it automatically gets 13 children which are the navigation children. By default is okay but it is very frustrating when there are many objects and you are trying to move the o...
importnumpyasnpres=np.array([])pd=inputs[0].PointData['S']forxx,yy,zz,xy,yz,xzinpd:t=np.array([[xx,xy,xz],[xy,yy,yz],[xz,yz,zz]])res=np.append(res,t)tensor=dsa.VTKArray(res)tensor.shape=(len(pd),3,3)output.PointData.append(tensor,'S_tensor') ...
百度试题 结果1 题目因式分解: x^2-xy+xz-yz=___.相关知识点: 试题来源: 解析 x^2-xy+xz-yz =(x^2-xy)+(xz-yz) =x(x-y)+z(x-y) = ( (x-y) ) ( (x+z) ). 故答案为: ( (x-y) ) ( (x+z) ).反馈 收藏