x y的范围是多少钱

由x2+xy+y2-2=0得：x2+2xy+y2-2-xy=0，即（x+y）2=2+xy≥0，所以xy≥-2；由x2+xy+y2-2=0得：x2-2xy+y2-2+3xy=0，即（x-y）2=2-3x

由题意可得a=x2−xy+y2x2+xy+y2=(xy)2−xy+1(xy)2+xy+1，令xy=t≠0，可得a=t2−t+1t2+t+1=1-2tt2+t+1=1-2t+1t+1，变形可得21−a−

因为x+y=ky=k-x代入1式x（k-x)=6-x^2+xk-6=0x^2-kx+6=0若方程有解k^2-4x6≥0则k≤-2√6或k≥2√6

x2+xy+y2=(x+y)2-xy=2,所以(x+y)2=2+xy.2|xy|+xy≤x2+xy+y2=2,所以0≤xy≤2/3.或者-2≤xy≤0u=x2-xy+y2=(x+y)2-3xy=2-2

将(a,-a)代入y^2=xy+2x+ka^2=-a^2+2a+kk=2a^2+2ak=2(a+1/2)^2-1/2k>=-1/2k的取值范围是(-1/2,正无穷)

设x^2-xy+y^2=p∵x^2+xy+y^2=1∴可知x^2+y^2=(1+p)/22xy=(1-p)∵x^2+y^2≥2|xy|∴(1+p)/2≥│(1-p)│∴1/3≤p≤3∴x^2-xy+y

可设z=x^2-xy+y^2.又1=x^2+xy+y^2.两式分别相加,减.得x^2+y^2=(z+1)/2.z-1=-2xy.两式再相加减得,(3z-1)/2=(x-y)^2≥0.（3-z)/2=(

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)+

将x+y=k变换得y=k-x然后代入第一个式子得x(k-x)=6化简得x^2-kx+6=0即为一元二次方程有实根的条件当然是det>=0了所以k^2-24>=0得k=2根号（6）

ax-2y=x+3ax-x-2y=3(a-1)x-2y=3∴a-1≠0a≠1

由基本不等式得x2+y2>=2根号(x2y2)=2丨xy丨,即2丨xy丨

x^2+xy+y^2=2(x-y)^2+3xy=2(x-y)^2=2-3xy≥0xy≤2/3x^2+xy+y^2=2(x+y)^2-xy=2(x-y)^2=2+xy≥0xy≥-2所以-2≤xy≤2/3

设x^2+y^2-xy＝t(1)x^2+y^2+xy＝1(2)由(1)(2)可解得：x^2+y^2=(t+1)/2(3)2xy=1-t(4)(3)+(4)化简得：(x+y)^2=(3-t)/2(3)-

楼上的不对,因为取-1的条件是x=y=0,而此时x^2+xy+y^2=1不成立.x^2-xy+y^2＝x^2+xy+y^2-2xy=1-2xyx^2+xy+y^2=1≥3xyxy≤1/3-2xy≥-2

因为X平方,y平方一定大于等于0将等式变换为：x平方+y平方=1-xy可得：xy=0所以：xy>=-1综上所述可得：-1

由题得xy=(1-x2)/2,因为x2>=0,所以xy=(1-x2)/2

(x+y)^2=1+3xy(x-y)^2=1-xyu=(x+y)(x-y)|u|=√(x+y)^2√(x-y)^2=√(1+3xy)√(1-xy)=√[-3(t-1/3)^2+2/3]≤√6/3故-√

xy+1=4x+y①∵x>0,y>0根据均值定理∴4x+y≥2√（4x*y)=4√(xy)②①②==>xy+1≥4√(xy)∴(xy)-4√(xy)+1≥0解得√(xy)≥2+√3或0

2x-3y=a-1x+2y=a解得x=(5a-2)/7>0a>2/5y=(a+1)/7>0a>-1所以a>2/5