已知4X^2-2X-1=0,求(12X^5-6X^4-7X^3+2X^2+1)/(2X^3-X+1)的值
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已知4X^2-2X-1=0,求(12X^5-6X^4-7X^3+2X^2+1)/(2X^3-X+1)的值
![已知4X^2-2X-1=0,求(12X^5-6X^4-7X^3+2X^2+1)/(2X^3-X+1)的值](/uploads/image/z/10857789-45-9.jpg?t=%E5%B7%B2%E7%9F%A54X%5E2-2X-1%3D0%2C%E6%B1%82%2812X%5E5-6X%5E4-7X%5E3%2B2X%5E2%2B1%29%2F%282X%5E3-X%2B1%29%E7%9A%84%E5%80%BC)
lijuan8687833 :
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
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
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