如果x y=-1,x-y=-3,则x的平方-y的平方是多少

(x^2+xy+2y^2)/(x^2-xy+y^2)=[(x/y)²+(x/y)+2]/[(x/y)²-(x/y)+1]分子分母同时除以y²=(4/9+2/3+2)/(4

x+y=3xy可得(x+y)/xy=3,得1/y+1/x=3答案就是3

x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7

(x-y)^2=x^2-2xy+y^2=9x^2+2xy+y^2=9+4=13(x+y)^2=13

1.-3x+4y原理：(a+b)·(a-b)=a²-b²本题中a=-3xb=4y2.x²+y²=(x+y)²-2xy=6²-2*7=36-1

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15

原式=3x-3y-6xy=3-12=-9

x²y-xy²=xy(x-y)=3×2=6再问：怎样变为xy(x-y)再答：提取公因式x²y＝(xy)×(x)xy²＝(xy)×(y)都有因式xy，可以提取出来

3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*

如果绝对值X-Y+1+根号（X-3）的平方=0x-y+1=0x-3=0x=3y=4xy=3x4=12

（1）（-2）*4=(-2)x4+1=-7（2）（-1*3）*（2）=[(-1)x3+1]*2=(-2)*2=(-2)x2+1=13（3）任意选择两个有理数,分别填入下列□和○内,并比较两个运算结果,

题目是这样吧1=xy/(x+y),2=yz/(y+z),3=xz/(x+z)倒数法,写成每个式子的倒数；1=1/x+1/y,(1)1/2=1/y+1/z,(2)1/3=1/x+1/z(3)三式相加,得

（x+2y-3xy)-(-2x-y+xy)=x+2y-3xy+2x+y-xy=(1+2)x+(2+1)y-(3+1)xy=3x+3y-4xy=3(x+y)-4xy=3*1/2-4*(-1/2)=3/2

1/x-1/y=2(y-x)/xy=2y-x=2xy(3x+5xy-3y)/(x-xy-y)=[5xy-3(y-x)]/[-xy-(y-x)]=(5xy-6xy)/(-xy-2xy)=-xy/-3xy

因为xy/（x+y）=1/2所以x+y=2xy原式=3（x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

2x+3y-3xy+4xy-3x-4y+7=(2x-3x+3y-4y)+4xy-3xy+7=(-x-y)+xy+7=-(x+y)+xy+7=-4-1+7=2

x：y=2:3设x=2k,则y=3k∴x²+xy+2y²/x²-xy+y²=(4k²+6k²+9k²)/(4k²-6k&

已知xy+x+y=6则xy+x+y+1=（x+1）（y+1）=7=7×1因为x,y都是自然数（非负整数）所以x+1,y+1≥1所以只能x+1=7y+1=1即x=6,y=0或x+1=1,y+1=7即x=

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采