已知多项式A=2x2-3xy 2y2,B=2x2 xy-3y2
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多项式3x2-34x4y-1.3+2xy2有4项组成,最高项是-34x4y,次数是5,常数项是-1.3.∴(1)四项式;(2)3x2,-34x4y,-1.3,2xy2;(3)-34x4y;(4)5次;
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
①x2y+xy2=xy(x+y)=1×3=3;②x2+y2=(x+y)2-2xy=32-2×1=7.
-2x2-12xy2+8xy3=-2x(x+6y2-4y3),故公因式是-2x.故答案为:-2x.
原式=(4x2y+5xy2+3x-2y+5)-2(2x2y-3xy2-2x+1)=4x2y+5xy2+3x-2y+5-4x2y+6xy2+4x-2=11xy2+7x-2y+3.
已知x=2-√3,y=2+√3,求代数式x2次方+y2次方值xy=4-3=1x+y=4x²+y²=(x+y)²-2xy=16-2=14再问:谢谢你,但我要那种初级复杂的过
A=B+C=(2x2-3xy-y2)+(x2+xy+y2)=2x2-3xy-y2+x2+xy+y2=3x2-2xy.故选D.
(1)原式=-(a-1)2;(2)原式=-x(x-y)2;(3)原式=(x+1)2(x-1)2;(4)原式=(x+y-x+y)2=4y2.
是不是求:5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问:是再答:已知是不是(x+3)²+|x+y+10|=
3xy(x²y-xy²+xy)-xy²(2x²-3xy+2x)=3x³y²-3x²y³+3x²y²-
(1)-X2(2)-3ab-9bc+6ac(你打错了吧,应该是2ab-3bc-ac吧)(3)-4
A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2
由于多项式A=x2+2y2-z2,B=-4x2+3y2+2z2且A+B+C=0,则C=-A-B=-(x2+2y2-z2)-(-4x2+3y2+2z2)=-x2-2y2+z2+4x2-3y2-2z2=3
原式=x3+3x2y-5xy2+6x3+1-2x3+y3+2xy2+x2y+2-4x2y-7x3-y3+4xy2+1=-2x3+xy2+4,由于y为偶次幂,故误把“x=3,y=-1”写成“x=3,y=
(2x2y-xy2)-(x2y-3xy2)=2x2y-xy2-x2y+3xy2=x2y+2xy2.故选C.
根据题意:A=8x3-7x2+5x+(3x3-2x2-4)=8x3-7x2+5x+3x3-2x2-4=11x3-9x2+5x-4.
A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²
mx3+3nxy2+2x2-xy2+y=mx3+(3n-1)xy2+2x2+y不含三次项式所以m=03n-1=0n=1/3因而2m+3n=1不懂再问哦
(-2x2)(3x2-ax-6)-3x3+x2=-6x^4+2ax³+12x²-3x³+x²=-6x^4+(2a-3)x³+13x²∵不含x
由题意得:3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy,∴C=13x2y+xy2−5xy3,故:C-A=13x2y+xy2−5xy3-(8x2y-6