已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是_____________.
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已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是_____________.
已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是
已知x≥1
x-y+1≤0
2x-y-2≤0
则x2+y2的最小值是_____________.
已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是已知x≥1 x-y+1≤0 2x-y-2≤0 则x2+y2的最小值是_____________.
x-y+1≤0 2x-y-2≤0
有Y≤-1-X,Y≤2-2X
x≥1有-X≤-1
则Y≤-2且Y≤0,所以Y≤-2.那么Y^2≥4
x≥1.所以X^2≥1
所以X^2+Y^2≥1+4=5
即最小制是5
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