$a)M=(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3) : \dfrac{2-4x}{x+1} - \dfrac{3x-x^2+1}{3x} \; (\textrm{ĐKXĐ} : x \ne 0; x\ne -1 ) \\
=[\dfrac{(x+2)(x+1)}{3x(x+1)}+\dfrac{6x}{3x(x+1)}-\dfrac{9x(x+1)}{3x(x+1)}].\dfrac{x+1}{2-4x} - \dfrac{3x-x^2+1}{3x} \\
=[\dfrac{x^2+3x+2}{3x(x+1)}+\dfrac{6x}{3x(x+1)}-\dfrac{9x^2+9x}{3x(x+1)}].\dfrac{x+1}{2-4x} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{x^2+3x+2+6x-(9x^2+9x)}{3x(x+1)}.\dfrac{x+1}{2(1-2x)} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{-8x^2+2}{3x(x+1)}.\dfrac{x+1}{-2(2x-1)} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{-2(4x^2-1)}{3x(x+1)}.\dfrac{x+1}{-2(2x-1)} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{-2(2x-1)(2x+1)}{3x(x+1)}.\dfrac{x+1}{-2(2x-1)} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{-2(2x-1)(2x+1)(x+1)}{3x(x+1).[-2(2x-1)]} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{2x+1}{3x} - \dfrac{3x-x^2+1}{3x} \\
=\dfrac{2x+1-(3x-x^2+1)}{3x} \\
=\dfrac{x^2-x}{3x} \\
=\dfrac{x(x-1)}{3x} \\
=\dfrac{x-1}{3} \\~\\
b) \textrm{Thế } x=2013 \text{ vào biểu thức trên ta được : } \\
\dfrac{x-1}3 \\
=\dfrac{2013-1}3 \\
=\dfrac{2012}3 \\~\\
c) \textrm{Để } M<0 \\
\textrm{Thì} \dfrac{x-1}3 < 0 \\
\iff x-1 < 0 \\
\iff x < 1 \\~\\
d) \dfrac1{M}=\dfrac{3}{x-1} \\
\textrm{Để } \dfrac1{M} \in Z \\
\iff 3 \quad \vdots \quad x-1 \\
\iff x-1 \in Ư\{3\}=\{ -3;-1;1;3 \} \\
\iff x = \{ -2; 0 (\textrm{loại}) ; 2 ; 4 \}$
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