用配方法证明2x^2-x+3的值恒大于0

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2024-11-30 04:56:51
推荐回答(3个)
回答(1):

2x^2-x+3
=2(x^2-x/2)+3
=2[x^2-(1/2)x+(1/4)^2-(1/4)^2]+3
=2[x^2-(1/2)x+(1/4)^2]-2×(1/4)^2+3
=2(x-x/4)^2+23/8
2(x-x/4)^2>=0
所以2(x-x/4)^2+23/8>=23/8>0
所以2x^2-x+3的值恒大于0

回答(2):

2x^2-x+3
=2(x^2-1/2x)+3
=2(x^2-1/2x+1/16-1/16)+3
=2(x-1/4)^2-1/8+3
=2(x-1/4)^2+23/8
(x-1/4)^2的值恒大于0
2x^2-x+3的值恒大于0

回答(3):

2x^2-x+3
=2(x^2-x/2)+3
=2(x^2-x/2+1/16-1/16)+3
=2(x-1/4)^2-1/8+3
=2(x-1/4)^2+23/8

∵(x-1/4)^2≥0
∴2(x-1/4)^2+23/8≥23/8>0