高一数学题，1.若3x^2+2y
解：1、3x²+2y²=2x ===> 2(x²+y²)=2x-x²=1-(x-1)²≥0
3(x-1/3)²+2y²=4/3
(x-1/3)²/(4/9)+y²/(2/3)=1
x∈[-2/3,2/3], y∈[-√6/3,√6/3]
设x=2cost/3+1/3, y=√6·sint/3
所以x²+y²=(2cost/3+1/3)²+(√6·sint/3)²
`````````=4cos²t/9+6sin²t/9+4cost/9...全部
解：1、3x²+2y²=2x ===> 2(x²+y²)=2x-x²=1-(x-1)²≥0
3(x-1/3)²+2y²=4/3
(x-1/3)²/(4/9)+y²/(2/3)=1
x∈[-2/3,2/3], y∈[-√6/3,√6/3]
设x=2cost/3+1/3, y=√6·sint/3
所以x²+y²=(2cost/3+1/3)²+(√6·sint/3)²
`````````=4cos²t/9+6sin²t/9+4cost/9+1/9
`````````=4/9+2sin²t/9+4cost/9+1/9
`````````=7/9-2cos²t/9+4cost/9
`````````=-(2/9)(cost-1)²+1
所以x²+y²∈[1/9,1]
2、f(x)=x²+ax+3-a=(x+a/2)²-a²/4-a+3在闭区间[-2,2]恒取非负数
(1)对称轴x=-a/2∈[-2,2],a∈[-4,4]时,只要f(-a/2)≥0即可
即a²/4+a-3≤0 ===> a²+4a-12=(a+6)(a-2)≤0,-6≤a≤2
故此时,-4≤a≤2。
(2)对称轴x=-a/2∈(2,+∞),a∈(-∞,-4)时,只要f(2)≥0即可
即7+a≥0 ===> a≥-7 ===> -4＞a≥-7
(3)对称轴x=-a/2∈(-∞,-2),a∈(4,+∞)时,只要f(-2)≥0即可
即7-3a≥0 ===> a≤7/3 ===> 无解
综合(1)、(2)、(3)a∈[-4,-2]∪[-7,-4)=[-7,2]。
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