1 2x-1 x y(x y 2x-x-y)
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(x+1)(y+1)
(x²+xy-12)²+(xy-2y²-1)²=0由于平方数都大于或等于0,所以上式成立的前提是:(x²+xy-12)²=0,即:x&sup
1.x^4+5x^2-6=(x^2+6)(x^2-1)=(x^2+6)(x-1)(x+1)2.(x+y)(x+y+2xy)+(xy+1)(xy-1)=(x+y)^2+2xy(x+y)+(xy+1)(x
根据题意,分式xy2x+3y中的x,y都扩大5倍,∴5x•5y2×5x+3×5y=25xy10x+15y=5xy2x+3y;即分式的值是原式值的5倍.故选B.
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
3xy[2xy-x(y-2)+x-1]=3xy(2xy-xy+2x+x-1)=3xy(xy+3x-1)=3x^2y^2+9x^2y-3xy
18(4x²-xy)-19(4x²-xy)+(4x²-xy-1)=72x²-18xy-76x²+19xy+4x²-xy-1=-4x²
这是要做什么?简化算式?分解因式?还是要怎样?
xy-1+x-y=XY+X-Y-1(加法交换律)=(XY+X)-(Y+1)=X(Y+1)-(Y+1)(提取公因式X)=(Y+1)(X-1)(提取公因式Y+1)这样可以么?
xy+x/x的平方+2x+1=x(y+1)/(x+1)^2不能继续约分了题目如果是xy+y/x的平方+2x+1=y(x+1)/(x+1)^2=y/(x+1)
1-x-y+xy=(1-x)-(y-xy)=(1-x)-y(1-x)=(1-x)(1-y)
原式=(xy)2+2xy+3-2(x+y)-1-(x+y)2+2(x+y)-1,=(xy)2+2xy+1-(x+y)2,=(xy+1)2-(x+y)2,=(xy+1+x+y)(xy+1-x-y),=(
1、=3x(1-4x²)=3x(1+2x)(1-2x)2、=(x²-2xy+y²)-1=(x-y)²-1=(x-y+1)(x-y-1)
原式=1-3xy+x+(-4x-6xy)=1-3xy+x-4x-6xy=1-3x-9xy
xy-x-y+1=x(y-1)-y+1=x(y-1)-(y-1)=(y-1)(x-1)
1.12x2+4x-1=12(x+1/2)(x-1/6)2.x2-(a+b)xy+aby2=(x-ay)(x-by)3.xy-1+x-y=(x-1)(1+y)4.x2-(2a-1)x+a2-a=(x-
因式分解x²(x+1)-y(xy+x)原式=x³+x²-xy²-xy=x³-xy²+x²-xy=x(x²-y²
(3x^2+2xy-1/2x)-(2x^2-xy+x)=3x^2+2xy-1/2x-2x^2+xy-x=3x^2-2x^2+2xy+xy-1/2x-x=x^2+3xy-3/2x(1/2xy+y^2+1
(x-1)(y-1)
解:原式=xy+y-x-1=y(x+1)-(x+1)=(x+1)(y-1)