大括号5x 2y等于15,8x 3y等于﹣1

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大括号5x 2y等于15,8x 3y等于﹣1
(x3-2y3-3x2y)-(3x3-3y3-7x2y)

原式=x3-2y3-3x2y-3x3+3y3+7x2y=-2x3+y3+4x2y

单项式:5x2y,-6x2y,34x

5x2y+(-6x2y)+34x2y=14x2y答:和是-14x2y.

已知A=x3-2y3+3x2y+xy2-3xy+4,B=y3-x3-4x2y-3xy-3xy2+3,C=y3+x2y+2

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.

一个多项式加上2x2y-3xy2-2x+1的2倍等于4x2y+5xy2+3x-2y+5,求这个多项式.

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.

当x=2011,y=2012时,求代数式3x3-4x3y2+3x2y+2x2+4x3y2+2x2y-5x2-5x2y+x

化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3

x4+y2x2+y4 x3+x2y-xy2-y3 (x2+x)-8(x2+x)+12 因式分解,x2表示x的两次

x4+y2x2+y4=x^4+2y^2x^2+y^4-x^2y^2=(x^2+y^2)^2--x^2y^2=(x^2+y^2+xy)(x^2+y^2-xy)x3+x2y-xy2-y3=(x-y)(x^

有这样一道题:“计算(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)的值,其中x=1

(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=-2y3=-2×(-1)3=2.因为化简的

已知A=x3+3x2y-5xy2+6y3-1,B=y3+2xy2+x2y-2x3+2,C=x3-4x2y+3xy2-7y

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

若x+y=-1,则x4+5x3y+x2y+8x2y2+xy2+5xy3+y4的值等于(  )

原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2

一个多项式加上3x2y-3xy2得x3-3x2y,则这个多项式是(  )

根据题意得:(x3-3x2y)-(3x2y-3xy2)=x3-3x2y-3x2y+3xy2=x3-6x2y+3xy2,故选C.

小明在解答题目:“已知x=3,y=-1,求代数式(x3+3x2y-5xy2+6x3+1)-(2x3-y3-2xy2-x2

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=

一个多项式加上-2x3-3x2y+5y2,得x3-2x2y+3y2.

(1)(x3-2x2y+3y2)-(-2x3-3x2y+5y2)=x3-2x2y+3y2+2x3+3x2y-5y2=3x3+x2y-2y2,答:这个多项式为3x3+x2y-2y2.(2)当x=-12,

4x2y-{x2y-「3xy2 – 1/ 2(4x2y-8xy2)+x2y」}-5xy2

答案:2x^2y+2xy^2原式=4x2y-{x2y-[3xy2-2x2y+4xy2+x2y]}-5xy2=4x2y-{x2y-[7xy2-x2y]}-5xy2=4x2y-{x2y-7xy+x2y]}

把多项式3x2y-5xy3+y2-2x3按x的降幂排列是 ___ .

多项式3x2y-5xy3+y2-2x3的各项为3x2y,-5xy3,y2,-2x3,按x的降幂排列为-2x3+3x2y-5xy3+y2.故答案为:-2x3+3x2y-5xy3+y2.再问:为什么是这样

把多项式3x2y-4xy2+x3-5y3按y的降幂排列是______.

由题意得:多项式3x2y-4xy2+x3-5y3按y的降幂排列是-5y3-4xy2+3x2y+x3.故答案是:-5y3-4xy2+3x2y+x3.

因式分解:x3-y3-x2y+xy2

x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)