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
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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
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=0
x=-1/2,y=4
2x+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
所以(x+y-1)^2+1>=1
即代数式(x+y)^2-2x-2y+2的值都不会小於1
4x^2+4x+y^2-8y+17=0
(4x^2+4x+1)+(y^2-8y+16)=0
(2x+1)^2+(y-4)^2=0
平方相加为0则都等于0
所以2x+1=0,y-4=0
2x=-1,y=4
所以2x+y=3
原式=(x+y)^2-2(x+y)+1+1
=(x+y-1)^2+1
平方≥0
所以(x+y-1)^2+1≥1
所以值都不会小於1
4x^2+4x+y^2-8y+17=0
则 (2x+1)^2+(y-4)^2=0
所以2x+1=0,y-4=0
所以2x+y=3
下面的把y=3-2x带入代数式化为只含有x的代数式
即x^2-4x+5=(x-2)^2+1>=1
所以无论x,y为何实数,代数式(x+y)^2-2x-2y+2的值都不会小於1
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