代数式10 |x 3| (2x 2y)的平方

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/01 03:29:46
代数式10 |x 3| (2x 2y)的平方
(x3-2y3-3x2y)-(3x3-3y3-7x2y)

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

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.

已知x+y=10,xy=24,求x3+y3-x2y-xy2的值

x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10

若代数式x3+y3+3x2y+axy2含有因式x-y,则a=______,在实数范围内将这个代数式分解因式,得x3+y3

∵代数式x3+y3+3x2y+axy2含有因式x-y,∴当x=y时,x3+y3+3x2y+axy2=0,∴令x=y,即x3+x3+3x3+ax3=0,则有5+a=0,解得a=-5.将a=-5代入x3+

x+y=5,xy=2,求代数式-x2y-xy2的值

解-x²y-xy²=-xy(x+y)=-2×5=-10

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

已知(x+3)2+▕x-y+10▏=0求代数式5x2y-【2x2-(3xy-xy2)-3x2】-2xy2-y2的值.

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|=

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则代数式x4+6x3y—2x2y+10x2y2—2xy2+6xy3+y4的值等于_____

原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²

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,

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

当x=-1,y=1时求代数式2x2y-(5xy2-3x2y)-x2的值

代入x=-1,y=1,2x^y-(5xy^-3x^y)-x^=2*(-1)^*1-{5*(-1)*1^-3*(-1)^*1}-(-1)^=2-(-5-3)-1=9备注:2^表示2的平方

当x等于1,y等于2时,求代数式3xy2-【2xy2-2(xy-1.5x2y)】+xy-3x2y的值

解3xy²-[2xy²-2(xy-1.5x²y)]+xy-3x²y=3xy²-(2xy²-2xy+3x²y)+xy-3x²

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)

已知x2-y2=xy,且xy≠0,求代数式x2y-2+x-2y2的值.

∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a