Question

设3x3－x＝1，求9x的四次方+12x3-3x2-7x+2001的值。

Answer 1

解：因为3x^3－x＝1
所以，原式=3x(3x^3-x)+12x^3-7x+2001
=3x+12x^3-7x+2001=12x^3-4x+2001=4(3x^3-x)+2001=4+2001=2005
此题一步步整体代入，代换求值的。

Answer 2

因为3x^3-x=1
所以3x^3-x-1=0
9x^4+12x^3-3x^2-7x+2001
=3x(3x^3-x-1)+12x^3-4x+2001
=4(3x^3-x-1)+4+2001
=2005

Answer 3

原题为3x^3-x=1,求9x^4+12x^3-3x^2-7x+2001.
由已知得，
x(3x^2-1)=1,又x≠0，所以
3x^2-1=1/x.
原式=9x^4-3x^2+12x^3-7x+2001
=3x^2(3x^2-1)+12x^3-7x+2001,(3x^2-1=1/x),
=3x+12x^3-7x+2001
=12x^3-4x+2001
=4x(3x^2-1)+2001
=4+2001
=2005.

Answer 4

3x3-x=1……可以得出x=8；
(9x)^4+12x3-3x2-7x+2001=72^4+36-6-56+2001
=26873856+30+1945
=26875831

Answer 5

3x3－x＝1 ∴ 3x4-x2=x
∴9x4-3x2=3x
9x4+12x3-3x2-7x+2001
=9x4-3x2 -7x ++12x3+2001
=12x3-4x+2001
=4（3x3－x）+2001
=2005