xy=( x-7)*(y 2)
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原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=(1x-y-yx-y)•xyy-1=1-yx-y•xyy-1=-xyx-y.故答案是:-xyx-y.
设x2-xy+y2=Ax2-xy+y2=A与x2+xy+y2=1相加可以得到:2(x2+y2)=1+A(1)x2-xy+y2=A与x2+xy+y2=1相减得到:2xy=1-A(2)(1)+(2)×2得
由x2+3xy+y2 x2+y2有意义,可知y与x不能同时为0.不妨设y≠0,由x2+xy-2y2=0,化为(x+2y)(x-y)=0,解得x=y,或x=-2y.把x=y代入,可得x2+3x
设x^2-xy+y^2=Ax^2-xy+y^2=A与x2+xy+y2=3相加可以得到2(x^2+y^2)=3+A(1)x^2-xy+y^2=A与x2+xy+y2=3相减得到2xy=3-A(2)(1)+
∵x<y<0,∴x-y<0,x+y<0.∴x2−2xy+y2=(x−y)2=|x-y|=y-x.x2+2xy+y2=(x+y)2=|x+y|=-x-y.∴x2−2xy+y2+x2+2xy+y2=-2x
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
由xy=0,得x=0,或y=0当x=0时,代入方程1:-y^2+根号y^2=a,即y^2-|y|+a=0,解得|y|=[1±√(1-4a)]/2当y=0时,代入方程1:x^2+根号x^2=a,即x^2
x^2+xy-2y^2-x+7y-6=(x+2y)(x-y)-x+7y-6x+2y-3Xx-y2十字相乘得:2(x+2y)-3(x-y)=-x+7y是一次项所以原式=(x+2y-3)(x-y+2)
∵2x2-xy-6y2=(x-2y)(2x+3y),∴可设2x2-xy-6y2+7x+7y+3=(x-2y+a)(2x+3y+b)a、b为待定系数,∴2a+b=7,3a-2b=7,ab=3,解得a=3
10拆成1+9X2-2X+1+Y2-6Y+9=0(X-1)2+(Y-3)2=0平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以X-1=0,Y-3=0X=1,Y=
分解因式有(x-3y)(2x-y)=0所以有x=3y或2x=y所以x:y=3:1或x:y=1:2
x²-2x+y²+6y+10=0,变换得(x-1)²+(y+3)²=0,∴x=1,y=-3∴(x2-2xy)/(xy+y2)=(1²-2*(-3))/
X2次方+Y2次方=(X+Y)^2-2XY=7^2-2*2=45再问:这个^是什么再答:x^2表示x的2次方
x2+y2=(x+y)2-2xy=(5)2-2×7=25-14=11
∵x^2+y^2=10,x+xy+y=7∴7-xy=x+y,且2xy≤x^2+y^2=10,∴(7-xy)^2=(x+y)^2=x^2+y^2+2xy=10+2xy,且xy≤5∴49-14xy+(xy
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
x^2-y^2=2xy,得x/y-y/x=2,即(y/x)^2+2(y/x)-1=0∴y/x=-1+√2或y/x=-1-√2(舍去,因为x,y都是正数).即(x-y)/(x+y)=√2-1
由x2+xy+y2=3得,x^2+y^2=3-xyx^2+y^2≥2xy得,xy≤1所以x^2-xy+y^2=3-2xy≥1等号成立当且仅当x=y=±1
7(x+y)=3(x^2-xy+y^2)=3[(x+y)^2-3xy]必有x+y=3k,k为整数因此7k=9k^2-3xy故也有k=3n-->x+y=9n,n为整数7n=27n^2-xy-->xy=2
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²