4x^2-12x+36y^2+60y+35

来源：百度知道编辑：UC知道时间：2024/05/31 14:43:43

验证无论xy为何值，4x^2-12x+36y^2+60y+35的值恒为正

4x^2-12x+36y^2+60y+35
=(2x)^2-12x+3^2 +(6y)^2+60y+5^2+1
=(2x-3)^2+(6y-5)^2+1

(2x-3)^2≥0 +(6y-5)^2≥0

4x^2-12x+36y^2+60y+35≥1

4x^2-12x+36y^2+60y
=(4x^2-12x+9)+(36y^2+60y+25)+1
=(2x-3)^2+(6x+5)^2+1>=1>0
所以恒为正

因为,4x^2-12x+36y^2+60y+35=(2x-3)^2+(6y+5)^2+1>=1>0

所以,4x^2-12x+36y^2+60y+35恒为正
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回答完毕...

4x^2-12x+36y^2+60y+35
＝（2x-3)^2+(6y+5)^2+1
>=1

=(2x-3)^2+(6y+5)^2+1

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