2. 1<=x^2+y^2<=2 求证：1/2<=x^2+xy+y^2<=3

来源：百度知道编辑：UC知道时间：2024/05/12 02:42:15

x=asinb,y=acosb
1<=x^2+y^2<=2,1<=a^2<=2
x^2+xy+y^2
=a^2+a^2sinbcosb
=a^2[1+1/2*sin2b]
由于-1<=sin2b<=1
那么
1/2*1=0.5<=1/2a^2<=a^2[1+1/2*sin2b]<=3/2a^2<=3/2*2=3
即
0.5<=x^2+xy+y^2<=3

1<=x^2+y^2<=2 推出 <=xy<= 推出1/2<=x^2+xy+y^2<=3

都算了~~

x^2+y^2>=2xy所以2xy<=2,xy<=1 x^2+y^2>=-2xy,2xy>=-(x^2+y^2)
而-(x^2+y^2)>=-2,xy>=-1
所以0<=x^2+xy+y^2<=3，题目好像打错了吧

已知cos(x-y/2)=-1/9,sin(x/2-y)=2/3,0<x<pi(圆周率),0<y<pi/2.求 cos(x+y)的值 y=x^2(1-x) (0<x<1) 求x的最值~ 2. 1<=x^2+y^2<=2 求证：1/2<=x^2+xy+y^2<=3 设0<x<1<y<2,则√x^2+y^2-2xy+4x-4y+4 + √1-2x+x^2 - √y^2-4y+4=? 函数y=x(1-x^2)(0<x<1)的最大值是求Y=X(1-X^2)的最值（0<X<1）求函数y=x^2-x^3 (0<x<1)的最大值 "函数y=x+x/1(-2<x<0)的极大值求z=2x+y的最大值，使y<=x,x+y<=1,y>=-1满足这些约束条件设x>0,y>0且x不等于y求证(x^3+y^3)1/3<(x^2=y^2)1/2