把多项式2x∧4-3x³y-x³+y²写成两个二项式的差,使被减式中不含字母y-搜答案-拍题作业网

3x^2y-3xy^2+y^3-x^3=y³-3xy²+3x²y-x³

原式=(x²-4y²)+x-2y=(x+2y)(x-2y)+(x-2y)=(x-2y)(x+2y+1)

1,(x-y)³-(x-y)²(x+y)=(x-y)²(x-y-x-y)=-2y(x-y)²2,x(x²-xy)-(4x²-4xy)=x&#

=(x+y)(x-y)-2x-4y-3=(x+y)(x-y)+(x-y)-3x-3y-3=(x-y)(x+y+1)-3(x+y+1)=(x-y-3)(x+y+1)

-1/2y^4-5x^2y^3+x^3y^2+2x^4y+1

3m(x-y）^2-9n(y-x)^3=3(x-y)^2(m-3n(y-x))=3(x-y)^2*(m+3nx-3ny)另一个多项式为:m+3nx-3ny

差是：(3x+2y)-(4x-2y)=3x+2y-4x+2y=-x+4y

啊

-2x²-2x+xy+y²

x^2-y^2-2x-4y-3=x^2-2x+1-y^2-4y-4=(x^2-2x+1)-(y^2+4y+4)=(x-1)^2-(y+2)^2=(x-1+y+2)(x-1-y-2)=(x+y+1)(x

（3x-2y）-(4x-2y)=3x-2y-4x+2y=-x

4x^2-3xy-2+5y^2=(4x^2-2)-(3xy-5y^2)

x的降幕3x(4)y-3x(3)y(2)+2x(2)y(4)-y(3)xy的升幕3x(4)y-3x(3)y(2)-y(3)x+2x(2)y(4)

/>(1)原式=(x+2y)(x-2y)+x-2y=(x-2y)(x+2y+1)第二题狮子看不清,能在描述一遍吗再问：分解因式：（x²+1）²-x²+x（x-2）（x&#

3x-2y-(4x-2y)=3x-2y-4x+2y=-x

由题意得：差=3x+2y-（4x-2y），=-x+4y．故填：-x+4y．

x^3+4x^2y-3xy-y^3

x²-y²-2x-4y-3=(x＋y)(x－y)-2x-4y-3=(x＋y＋1)(x－y－3)(十字相乘)十字相乘x＋y1×x－y﹣3x^2y+y^2x=96xy(x＋y)=96x

因为,多项式2x^4-3x^2y-x^3+y^2中不含字母y的单项式为2x^4和x^3；所以,把多项式写成两个二项式的差,使被减式中不含字母y,结果如下：2x^4-3x^2y-x^3+y^2=(y^2

用提取公因式法,提取（x+3y)原式=（x+3y)[(3x-4y)-(x-2y)]=(x+3y)(2x-2y)=2(x+3y)(x-y)