(x加y)的二次方减6(x加y)加9因式分解-搜答案-智慧作业帮

x^2y^2-4xy+x^2+y^2+1=(xy)^2-2xy+1+x^2-2xy+y^2=(xy-1)+(x-y)^2=0xy-1=0,x-y=0,x^2=y^2=1,x=y=1或x=y=-1x+y

x=7,y=-1

由（1）x²-xy=5（2）y²+xy=-2∴（1）+（2）得：x²+y²=3.

x³-y³=(x-y)(x²+xy+y²)所以是乘x-y

先配方得（x+1）^2-(y-3)^2=0,再用平方差公式得（x+y-2）(x-y+4)=0.须知两个因数相乘得零的话这两个因数至少有一个为零.x+y不等于2即前一个因数不为零,即x-y+4=0,即x

X²+4X+Y²-6Y+13=0X²+4X+4+Y²-6Y+9=0（X+2）²+（Y-3）²=0X=-2,Y=3

X2-4X+Y2+6Y+13=0（X-2）2+（Y+3）2=0∴有X-2=0,X=2Y+3=0,Y=-3

x^2+y^2-4x-10y+29=0x^2-4x+4+y^2-10y+25=0(x-2)^2+(y-5)^2=0x=2y=5x^y+xy^2=xy(x+y)=2*5(2+5)=70

（x加y加z）的二次方减（x加y加z）的二次方=(x+y+z+x+y+z)(x+y+z-x-y-z)=0你抄错了吧再问：再答：（x加y加z）的二次方减（x加y-z）的二次方=(x+y+z+x+y-z)

x/y=3x=3y原式=[(3y)²+2*3y*y-3y²]/[(3y)²-3y*y+y²]=(9y²+6y²-3y²)/(9y&

已知x=二分之一,xy(y加y的二次方)减y的二次方(xy-x)+2x(x-y的平方)的值=xy²+xy³-xy³+xy²+2x²-2xy²

解题思路：考查多项式的运算解题过程：解：因为2x2+my-12与nx2-3y+6的和中不含x，y所以n=-2，m=3所以mn=-6祝你取得优异成绩！加油！最终答案：略

7x^3-6x^3y+3x^2y+3x^3+6x^3y-3x^2y-10x^3,其中x=-2010.y=2012=(7x^3+3x^3-10x^3)+(6x^3y-6x^3y)+(3x^2y-3x^2

x二次方加y的二次方减6x加4y加13的平方根等于零x^2+y^2-6x+4y+13=0x^2-6x+9+y^2+4y+4=0(x-3)^2+(y+2)^2=0x-3=0,y+2=0x=3,y=-2x

x=2y=1

原式=[x/(x-y)]*[y^2/(x+y)]-[x^4y/(x^2+y^2)(x^2-y^2)]/[x^2/(x^2+y^2)]=xy^2/(x^2-y^2)-x^2y/(x^2-y^2)=xy(

x³y³－2x²y²＋xy＝xy（x²y²－2xy＋1）＝xy（xy－1）²

再问：不清楚再答：字不清楚or看不懂再问：都再答：平方差公式啊，学了吧再问：嗯再答：再问：谢了

=x/2-(x-2y^2)+(-3x+y^2)=x/2-x+2y^2-3x+y^2=3y^2-7/2x