Giải thích các bước giải:
a.Ta có :
$x^{2k}-1\quad\vdots\quad x^2-1$
$x^{2k+1}-x=x(x^{2k}-1)\quad\vdots\quad x^2-1$
$\to f(x)=x^{100}+x^{99}+x^{98}+..+x+1$
$\to f(x)=(x^{100}+x^{98}+..+x^2)+(x^{99}+x^{97}+..+x^3)+x+1$
$\to f(x)=(x^{100}-1+x^{98}-1+..+x^2-1)+(x^{99}-x+x^{97}-x+..+x^3-x)+x+1+50x^2+49x$
$\to f(x)=(x^{100}-1+x^{98}-1+..+x^2-1)+(x^{99}-x+x^{97}-x+..+x^3-x)+1+50x^2+50x$
$\to f(x)=(x^{100}-1+x^{98}-1+..+x^2-1)+(x^{99}-x+x^{97}-x+..+x^3-x)+50(x^2-1)+50x+51$
$\to f(x)$ chia $x^2-1$ dư 50x+51
b.Ta có : $M=\dfrac{x^2-4x+2013}{x^2}$
$\to M=1-\dfrac{4}{x}+\dfrac{2013}{x^2}$
$\to M=2013(\dfrac{1}{x}-\dfrac{2}{2013})^2+\dfrac{2009}{2013}\ge \dfrac{2009}{2013}$
Dấu = xảy ra khi $\dfrac{1}{x}-\dfrac{2}{2013}=0\to x=\dfrac{2013}{2}$
