若3x^4+x^3-4x^2-17x+5除以x^2+x+1的商式是ax^2+bx+c,余式是dx+e,求(a+b+c)^d+e的值
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若3x^4+x^3-4x^2-17x+5除以x^2+x+1的商式是ax^2+bx+c,余式是dx+e,求(a+b+c)^d+e的值若3x^4+x^3-4x^2-17x+5除以x^2+x+1的商式是ax
若3x^4+x^3-4x^2-17x+5除以x^2+x+1的商式是ax^2+bx+c,余式是dx+e,求(a+b+c)^d+e的值
若3x^4+x^3-4x^2-17x+5除以x^2+x+1的商式是ax^2+bx+c,余式是dx+e,求(a+b+c)^d+e的值
若3x^4+x^3-4x^2-17x+5除以x^2+x+1的商式是ax^2+bx+c,余式是dx+e,求(a+b+c)^d+e的值
用大除法!
3x^2 -2x -5
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x^2+x+1/3x^4+x^3-4x^2-17x+5
3x^4+3x^3+3x^2
-------------------
-2x^3-7x^2-17x
-2x^3-2x^2-2x
-------------------
-5x^2-15x+5
-5x^2-5x-5
-------------------
-10x+10
所以3x^4+x^3-4x^2-17x+5=(x^2+x+1)(3x^2 -2x -5)-10x+10
a=3,b=-2,c=-5,d=-10,e=10
(a+b+c)^d+e=(-4)^(-10)+10
有点怪,题没错吧?
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