Question

A=1+2x^4，B=x^2+2x^3，试比较A与B的大小。

Answer 1

A=1+2x^4，B=x^2+2x^3
A－岁庆铅差蔽B=2x^4-2x^3-x^2+1
=(x-1)*2x^3-(x-1)*(x+1)
=(x-1)*[2x^3-x-1]
=(x-1)*[x^3-x+x^3-1]
=(x-1)[x(x^2-1)+(x-1)(x^2+x+1)]
=(x-1)[x(x-1)(x+1)+(x-1)(x^2+x+1)]
=(x-1)[(x-1)x(x+1)+(x-1)(x^2+x+1)]
=(x-1)^2*(2x^2+2x+1)
=(x-1)^2*(x^2+2x+1+x^2)
=(x-1)^2*[(x+1)^2+x^2]>乎好=0
所以A>=B

Answer 2

A－B=2x^4-2x^3-x^2+1
=(x-1)*2x^3-(x-1)*(x+1)
=(x-1)*[2x^3-x-1]
=(x-1)*[x^3-x+x^3-1]
=(x-1)[x(x^2-1)+(x-1)(x^2+x+1)]
=(x-1)[x(x-1)(x+1)+(x-1)(x^2+x+1)]
=(x-1)[(x-1)x(x+1)+(x-1)(x^2+x+1)]
=(x-1)^2*(2x^2+2x+1)
=(x-1)^2*(x^2+2x+1+x^2)
=(x-1)^2*[(x+1)^2+x^2]>0
所以A>B

Answer 3

当汪世冲X=正负1时，A=B

其他情况都是A大于B 比较两个数大小，可以困歼两数相返吵减，看结果大于0 小于0 或者等于0

Answer 4

令f=A-B=1+2x^4-x^2-2x^3=(x-1)^2*x^2+(x^2-1)^2>=0，则A>=B恒成立

Answer 5

A-B=2x^4+1-2x^3_x^2因式分解然后讨论