Cho đa thức Q(x) = A = ($\frac{x^{2} }{2}$ – $\frac{1}{2}$ . $x^{3}$ + $\frac{1}{2}$x)x – ( ($\frac{x^{5}}{3}$ – $\frac{1}{2}$ . $x^{4}$ + $x^{2}$ –

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1. Giải thích các bước giải:

   Ta có:

   $Q(x)=(\dfrac{x^2}{2}-\dfrac12x^3+\dfrac12x)x-(\dfrac{x^5}{3}-\dfrac12x^4+x^2-\dfrac{x^5}{3})$

   $\to Q(x)=(\dfrac{x^2}{2}-\dfrac12x^3+\dfrac12x)x-(\dfrac{x^5}{3}-\dfrac{x^5}{3}-\dfrac12x^4+x^2)$

   $\to Q(x)=(\dfrac{x^2}{2}-\dfrac12x^3+\dfrac12x)x-(0-\dfrac12x^4+x^2)$

   $\to Q(x)=(\dfrac{x^2}{2}-\dfrac12x^3+\dfrac12x)x-(-\dfrac12x^4+x^2)$

   $\to Q(x)=\dfrac{x^3}{2}-\dfrac12x^4+\dfrac12x^2+\dfrac12x^4-x^2$

   $\to Q(x)=-\dfrac{1}{2}x^4+\dfrac{1}{2}x^4+\dfrac{x^3}{2}+\dfrac{1}{2}x^2-x^2$

   $\to Q(x)=\dfrac{x^3-x^2}{2}$

   $\to  Q(x)=\dfrac{x^2(x-1)}{2}$

   Ta có $x\in Z\to x, x-1$ là $2$ số tự nhiên liên tiếp

   $\to x(x-1)\quad\vdots\quad 2$

   $\to x^2(x-1)\quad\vdots\quad 2$

   $\to \dfrac{x^2(x-1)}{2}\in Z,\quad\forall x\in Z$

   $\to đpcm$

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