x三角形Y=X-Y除以2

因为正数x,y满足x平方-y平方=2xy,所以(x^2-y^2)=2xy即1-(y/x)^2=2(y/x)(y/x)^2+2(y/x)-1=0y/x=[-2+√(4+4)]/2=-1+√2(因为x,y

设全集U=｛（x,y)|x,y属于R｝,集合M={(x,y)|(y+2)/(x-2)=1},则M={(x,y)|(y=x-4且x≠2}所以,M的补集为{(x,y)|(y≠x-4}∪{(2,-2)}集合

首先你要理解所给集合的元素代表什么.全集U为平面点集,M为两条射线（直线y=x+1除去点（2,3））,N表示平面内除去直线y=x+1以外的点.我想这样你应该能得出结果了吧?有问题继续问我.再问：能再说

原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x

x^2=x的平方‘/’为除以[(x+2y)^2-(x+y)(x-y)-5y^2]/(2x)=[(x^2+4xy+4y^2)-(x^2-y^2)-5y^2]/(2x)=[4xy]/(2x)=2y[(3x

分子分母同时除以Y令x比y=Z即使求Z=?(2Z+3)/（5z-3）=3/515Z-9=10Z+15Z=24/5

[(x+2y)(x-2y)-(x-2y)^2+8y(x+y)]/4x=[x^2-4y^2-x^2-4y^2+4xy+8xy+8y^2]/4x=12xy/4x=3y

2X-Y/X+Y=2X+Y/2X-Y=1/2原式＝[2(2X-Y)/X+Y]-[4(X+Y)/2X-Y]=2*2-4*(1/2)=4-2=2P.S.第一次为团队答题,好鸡东哦o(∩_∩)o有什么不懂的

/>2x+y=(4x-y-1)/2=(10+6x+2y)/4=(5+3x+y)/24x+2y=4x-y-1……1式3y=-1y=-1/34x+2y=5+3x+y……2式x=5-y=5+1/3=16/3

y=(2x+3)/(x-1)=[2(x-1)+2+3]/(x-1)=2+5/(x-1)x＜1,∴5/(x-1)＜0,∴y＜2或∵x＜1,x-1≠0,变换得y(x-1)=2x+3,即x=(3+y)/(y

解原式=2(x-y)/(x+y)-(x+y)/3(x-y)=6(x-y)(x-y)/[3(x+y)(x-y)]-(x+y)x+y)/[3(x-y)(x+y)]=5(x-y)(x-y)/[(x+y)(x

(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4

2x-y/x+3y=2得x+3y/2x-y=1/24x-2y/x+3y减4x+12y除以2x-y=2(2x-y)/x+3y-4(x+3y)/2x-y=2乘以2-4乖以1/2=4-2=2

【(x-y)^2+(x+y)(x-y)】除以2x=(x-y)*(x-y+x+y)/2x=(x-y)*2x/2x=x-y

[(2x—y)(2x+y)—y(y—6x)]/2x=（4x²-y²-y²+6xy）/2x=（4x²-2y²+6xy）/2x=2x-y²/x+

原式=[(x-y)(x-y)+(x+y)(x-y)]/2x=(x-y)(x-y+x+y)/2x=(x-y)·2x/2x=x-y=3-1.5=1.5

令x除以2=y除以3=z除以4=k则x=2k,y=3k,z=4kx+y+z除以x-2y+3z=(2k+3k+4k)除以(2k-6k+12k)=9k除以8k=8分之9

绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=03x-y=0(1)x+y=0(2)(1)+(2)4x=0x=0,代入(2)y=-x=0(x-y)/(xy)无意义,因此题目错了.

（x+y）*(x-y)=12除以第二个式子得x+y=6,就可以求出x=4,y=2,所以答案为2