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已知实数x、y满足2x+y=1,求u=x²+y²+4x-2y的取值范围
答案:1 悬赏:20 手机版
解决时间 2021-05-01 21:47
- 提问者网友：感性作祟
- 2021-05-01 14:05
最佳答案
- 五星知识达人网友：想偏头吻你
- 2021-05-01 14:44
解：U=x^2+y^2+4x-2y=(x+2)^2+(y-1)^2-5
=[1/2(x+2)]^2+[1/2(x+2)]^2+[1/2(x+2)]^2+[1/2(x+2)]^2+(y-1)^2-5
≥[1/2(x+2)+1/2(x+2)+1/2(x+2)+1/2(x+2)+(y-1)]^2/5-5
=(2x+y+3)^2/5-5
≥(1+3)^2/5-5
=-9/5
=[1/2(x+2)]^2+[1/2(x+2)]^2+[1/2(x+2)]^2+[1/2(x+2)]^2+(y-1)^2-5
≥[1/2(x+2)+1/2(x+2)+1/2(x+2)+1/2(x+2)+(y-1)]^2/5-5
=(2x+y+3)^2/5-5
≥(1+3)^2/5-5
=-9/5
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