证明;x4+y4+z4-2x2y2-2x2y2-2y2z2能被(x+y+z)整除
已知x+y+z=0,求x4+y4+z4-2x2y-2y2z2-2z2x2的值
有这样一道题,计算(2x4-4x3y-x2y2)-2(x4-2x3y-y3)+x2y2的值,其中x=0.25,y=-1;
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x
若x+y=1则代数式x4+6x3y—2x2y+10x2y2—2xy2+6xy3+y4的值等于_____
计算(1)(a-b+c-d)(c-a-d-b);(2)(x+2y)(x-2y)(x4-8x2y2+16y4).
若x+y=-1,则x4+5x3y+x2y+8x2y2+xy2+5xy3+y4的值等于( )
已知x+y+z=1 x2+y2+z2=2 x3+y3+z3=3 求x4+y4+z4=?
对于多项式x4-y4-4x3y-x2y2+3xy3按x得降幂排列按y得升幂排列
(2x2y-2xy2)-[(-3x2y2+3x2y)+(3x2y2-3xy2)],其中x=-1,y=2.
化简并求值:(2x2y-2xy2)-[(-3x2y2+3x2y)+(3x2y2-3xy2)],其中x=−12,y=2
化简求值(2X2-2y2)-3(X2y2+X2)+3(X2y2+y2),其中X=-1,y=2
分解因式:16x4-72x2y2+81y4.