1.原式=8x^3 × (-2Y^3) ÷ 16xy^2=-16x^3y^3 ÷ 16xy^2=-x^2y
2.原式=(2x+y)^2 - (3z)^2=4x^2 + 4xy + y^2 - 9z^2
(2x)^3×（-2Y^3)÷16XY^2 (2X+Y+3Z)(2X+Y_3Z)
1个回答
相关问题
-
(-2xy^2)^2×(-2x^2y)^3×(-x^2 z)÷(2x^3y^2)
-
为什么（y3/z+z3/x+x3/y）(yz+zx+xy)>=(y2+x2+z2)2
-
证明：（y+z-2x）3+（z+x-2y）3+（x+y-2z）3=3（y+z-2x）（z+x-2y）（x+y-2z）．
-
已知x+y+z=1,x2+y2+z2=2,x3+y3+z3=3,求xy(x+y)+yz(y+z)+zx(z+x)的值
-
已知A=2x^3-xy^z,B=y^3-z^2+xy^z,C=-x^3+2y^2-xy^z,其中,x=-1,y=1,z=
-
化简 [x^3(y^2-z^2)+y^3(z^2-x^2)+z^3(x^2-y^2)]/[x^3(y-z)+y^2(z-
-
x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyz
-
x+y+z=3,xy+yz+xz=-1,xyz=2,求 x^3+y^3+z^3 (2+x)(2+y)(2+z) 1/x+
-
方程组(xy+x)/(x+y+1)=2 (xz+2x)/(z+y+2)=3 (y+1)(z+2)/(z+y+3)=4
-
一.｛x+2y+3z=14 2x-y-z=-3 3x+y+z=8 二.｛3X-2Y-2Z=3 2X+2Y+3Z=12 X