Question

已知4X^2-2X-1=0,求(12X^5-6X^4-7X^3+2X^2+1)/(2X^3-X+1)的值

Answer 1

你好，lijuan8687833 ：

解：
12x^5-6x^4-7x^3+2x^2+1
=4x^2(3x^3-x+1)-2x(3x^3-x+1)-(3x^3-x+1)-4x^2+2x+2
=(3x^3-x+1)(4x^2-2x-1)-(4x^2-2x-1)+1
=0-0+1
=1

2x^3-x+1
=1/2(4x^3-2x+2)
=1/2[x(4x^2-2x-1)+2x^2-x+2]
=1/2[x(4x^2-2x-1)+1/2(4x^2-2x-1)+5/2]
=1/2[(4x^2-2x-1)(x+1/2)+5/2]
=(1/2)×(5/2)
=5/4

∴(12x^5-6x^4-7x^3+2x^2+1)/(2x^3-x+1)
=1÷(5/4)
=4/5

Answer 2

分子
12X^5-6X^4-7X^3+2X^2+1
=3x^3(4X^2-2X-1)-x(4X^2-2X-1)-x+1
=1-x
分母
2X^3-X+1
=1/2x(4X^2-2X-1)+1/4(4X^2-2X-1)+5/4
=5/4
根据4X^2-2X-1=0，x=(1+√5)/4或（1-√5)/4
带入
(12X^5-6X^4-7X^3+2X^2+1)/(2X^3-X+1)
=4(1-x)/5
=(3+√5)/5或(3-√5)/5

Answer 3

-2/3

Answer 4

4/5