若A={(x,y)|y=x*2 2x-1}B={(x,y)|y=3x 1

求A∩B就是求Y=X²-2与X=Y²-2的公共解两式相减得：Y-X=X²-Y²故：(X-Y)(X+Y)+(X-Y)=0故：(X-Y)(X+Y+1)=0故：X=Y

A、B这两个集合都可以看成两条线上的点的集合,A交B为空集即联立方程组2x+y=1.①a*2x+2y=a...②无解将①改为y=1-2x代入②中：（2a-4）x=a-2当a=2时,x有无穷多解,即A、

A={(x,y)|2x+y=1,x,y属于R}B={(x,y)|a²x+2y=a,x,y属于R}若A交B=空集,说明两直线2x+y=1,a²x+2y=a无交点即说明两直线平行,不重

loga((x-y)/2)=(logax+logay)/2=1/2*loga(xy)=loga√xy所以(x-y)/2=√xy两边平方x²-2xy+y²=4xyx²-6x

Dx^y+x^-y=2根号2===>(x^y+x^-y)^2=8===>x^2y+x^-2y+2=8===>x^2y+x^-2y=6(x^y-x^-y)^2=x^2y+x^-2y-2=6-2=4==>

应该是B.3/4

是不是-2A+B=2a所以-4x²+6xy-2y²+2x-4y+4x²-6xy+2y²+3x-y=5x-5y=5a+15=2aa=-3A=18-27+9+3-6

x^2=2x^2+2x-3x^2+2x-3=0(x+3)*(x-1)=0x1=-3,x2=1y1=9,y2=1故A交B={(-3,9),(1,1)}

(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=

c∵x+y≠0∴x≠-y∵3x-6y/x+y=0∴3x-6y=0∴x=2y∴x=2y且x≠-y

2X-Y=9,X+2Y=A,则X:Y=?4X-2Y=18.①X+2Y=A.②①+②---5X=18+A.③2X-Y=9.④2X+4Y=2A.⑤⑤-④---5Y=9+2A.⑥⑤：⑥得X：Y=（18+A）

先一个一个的展开括号项,再同项合并就行了啊[x(x-y)-y(x-y)+(x+y)(x-y)]÷2x=[xx-xy-(xy-yy)+x(x-y)+y(x-y)]÷2x=[xx-xy-xy+yy+xx-

9（x+y)^(2m)*(x-y)^(4n)*[-(x+y)^2]=-9(x+y)^(2m+2)*(x-y)^(4n)∴a=92m+2=104n=12-n∴m=4n=12/5

全部Y表示,连立就是x^2+ax+2=a/2-a^2x/2无解~就是x^2+(a+a^2/2)x+2-a/2=0德儿塔小于0就OK了

设A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²-3x-y,且B-2A=a所以a=B-2A=4x²-6xy+2y²-3x-y

若(x-y)/(x+y)=3那么x-y=3(x+y)x-y=3x+3y故x=-2y所以(x+2y)/(6x-7y)=(-2y+2y)/(6*(-2y)-7y)=0如果不懂,请Hi我,祝学习愉快!

y=2/(x-2)得到：x-2=2/y,x=2/y+2∵y

∣x-2a∣+（y+3）²=0因为∣x-2a∣≥0,（y+3）²≥0所以∣x-2a∣=0,（y+3）²=0所以x=2a,y=-3B-2A=(4x²-6xy+2y

3x+2y=1x-y=2解联立方程得x=1,y=-1A∩B={(x,y)|x=1,y=-1}B和C所代表的两直线平行,无交点(即空集)所以B∩C={Φ}A和D所代表的两直线重合,所以A∩D=A={(x

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