tìm x nguyên để B nguyên B = 4x^3 – 3x^2 + 2x – 83 / x – 3

0 bình luận về “tìm x nguyên để B nguyên B = 4x^3 – 3x^2 + 2x – 83 / x – 3”

1. Đáp án: $x \in {\rm{\{ }} – 1;1;2;4;5;7\} $

    

   Giải thích các bước giải:

   $\begin{array}{l}
   B = \frac{{4{x^3} – 3{x^2} + 2x – 83}}{{x – 3}}\left( {dkxd:x \ne 3} \right)\\
    = \frac{{4{x^3} – 12{x^2} + 9{x^2} – 27x + 29x – 87 + 4}}{{x – 3}}\\
    = \frac{{4{x^2}\left( {x – 3} \right) + 9x\left( {x – 3} \right) + 29\left( {x – 3} \right) + 4}}{{x – 3}}\\
    = \frac{{\left( {x – 3} \right)\left( {4{x^2} + 9x + 29} \right) + 4}}{{x – 3}}\\
    = 4{x^2} + 9x + 29 + \frac{4}{{x – 3}}\\
   Để:B \in Z\\
    \Rightarrow \frac{4}{{x – 3}} \in Z\\
    \Rightarrow \left( {x – 3} \right) \in Ư\left( 4 \right) = {\rm{\{ }} – 4; – 2; – 1;1;2;4\} \\
    \Rightarrow x \in {\rm{\{ }} – 1;1;2;4;5;7\} \left( {tmdk} \right)
   \end{array}$

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