C(x)=x^21-2004x^20+2004x^19+…+2004x-1. Tại x=-2003

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C(x)=x^21-2004x^20+2004x^19+…+2004x-1. Tại x=-2003

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Mary 1 tháng 2021-09-11T12:27:34+00:00 1 Answers 5 views 0

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    2021-09-11T12:28:55+00:00

    Giải thích các bước giải:

    Ta có:

    $C(x)=x^{21}-2004x^{20}+2004x^{19}+…+2004x-1$

    $\to xC(x)=x^{22}-2004x^{21}+2004x^{20}+…+2004x^2-x$

    $\to xC(x)-C(x)=x^{22}-2005x^{21}-4008x^{20}-2005x+1$

    $\to (x-1)C(x)=x^{22}-2005x^{21}-4008x^{20}-2005x+1$

    $\to C(x)\dfrac{x^{22}-2005x^{21}-4008x^{20}-2005x+1}{x-1}$

    Tại $x=-2003$ ta có:

    $\to C(x)\dfrac{(-2003)^{22}-2005\cdot (-2003)^{21}-4008\cdot (-2003)^{20}-2005\cdot (-2003)+1}{-2003-1}$

    $\to C(x)\dfrac{2003^{22}+2005\cdot 2003^{21}-4008\cdot 2003^{20}+2005\cdot2003+1}{-2004}$

     

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