4xy2-4x2y-y3

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若x+y=2,xy=-4,求x2y+xy2+1的值

(x+y)(xy)=x^2y+xy^2=-8原式=-7

4xy2-4x2y-y3(2和3是次方)

解题思路:提取公因式解题过程:提取公因式最终答案:略

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=6,xy=4,则x2y+xy2的值为______.

∵x+y=6,xy=4,∴x2y+xy2=xy(x+y)=4×6=24.故答案为:24.

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

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.

把多项式x3-4x2y+4xy2分解因式,结果为______.

x3-4x2y+4xy2,=x(x2-4xy+4y2),=x(x-2y)2.

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

如果2x+y=4,xy=3,那么2x2y+xy2的值为______.

∵2x+y=4,xy=3,∴2x2y+xy2=xy(2x+y)=3×4=12.故答案为:12

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

把多项式3xy2-3x2y-y3+x3按字母x降幂排列为______.

多项式3xy2-3x2y-y3+x3的各项为3xy2,-3x2y,-y3,x3,按x的降幂排列为x3-3x2y+3xy2-y3.

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]}

因式分解:(1)(a+b)x2-(a+b);(2)9(a+b)2-4(a-b)2;(3)6xy2-9x2y-y3;(4)

(1)(a+b)x2-(a+b)=(a+b)(x2-1)=(a+b)(x+1)(x-1);(2)9(a+b)2-4(a-b)2=[3(a+b)+2(a-b)][3(a+b)-2(a-b)]=(3a+3

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)

单项式5x2y,3x2y2,-4xy2的和为______.

5x2y+3x2y2+(-4xy2)=5x2y+3x2y2-4xy2.

已知x+2y=5,xy=1.则2x2y+4xy2=______.

∵x+2y=5,xy=1,∴2x2y+4xy2=2xy(x+2y)=2×1×5=10,故答案为:10.

已知x-y≠0 x2-x=7 y2-y=7 求x3+y3+x2y+xy2的值

x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup