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4x^2+4x+y^2-8y+17=0,求2x+y的值
试说明无论x,y为任何实数,代数式(x+y)^2-2x-2y+2的值都不会小於1
人气:495 ℃ 时间:2020-03-30 21:58:59
解答
4x^2+4x+y^2-8y+17=0,4x^2+4x+1+y^2-8y+16=0,(2x+1)^2+(y-4)^2=0(2x+1)^2=0,(y-4)^2=0x=-1/2,y=42x+y=(-1/2)*2+4=-1+4=3(x+y)^2-2x-2y+2=(x+y)^2-2(x+y)+2=(x+y)^2-2(x+y)+1+1=(x+y-1)^2+1因为(x+y-1)^2>=0所...
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