3x方+2y方=2x,求x方+y方范围

3x²+2y²=2x
可得 y²= （2x-3x²）/2 ≥0
即2x-3x²≥0
3x²-2x≤0,
x（3x-2）≤0,解得 0≤x≤2/3
x^2+y^2
=x²+（2x-3x²）/2
=-1/2 *x²+x= -1/2*(x-1)²+ 1/2
对称轴x=1,在0≤x≤2/3单调递增
故当x= 2/3 时,最大值是 4/9
当x=0时,最小值是 0
取值范围是[0,4/9]

3x^2-2x+2y^2=0
x^2-2x/3+2y^2/3=0
(x-1/3)^2+y^2/(3/2)=1/9
(x-1/3)^2/(1/9)+y^2/(1/6)=1
所以可以令x=1/3*cosa+1/3 y=1/√6*sina 0<=a<2π
x^2+y^2
=cos^2a/9+2cosa/9+1/9+sin^2a/6
=5/1...