$A =\dfrac{x^2}{x^3 - 4x} + \dfrac{6}{6 - 3x} +\dfrac{1}{x+2}$
a) $ĐKXĐ:\begin{cases}x^3 - 4x\ne 0\\6 - 3x \ne 0\\x + 2\ne 0\end{cases}\to \begin{cases}x \ne -2\\x \ne 0\\x \ne 2\end{cases}$
b) $A = \dfrac{x}{x^2 - 4} +\dfrac{2}{2 - x} + \dfrac{1}{x+2}$
$\to A =\dfrac{x}{(x-2)(x+2)} -\dfrac{2(x+2)}{(x-2)(x+2)} +\dfrac{x-2}{(x-2)(x+2)}$
$\to A =\dfrac{x- 2x - 4 + x - 2}{(x-2)(x+2)}$
$\to A =- \dfrac{6}{(x-2)(x+2)}$
c) Khi $x = 3$ ta được:
$A = -\dfrac{6}{(3-2)(3+2)}=-\dfrac65$
d) $A = 2$
$\to -\dfrac{6}{(x-2)(x+2)} = 2$
$\to x^2 - 4 = -3$
$\to x^2 = 1$
$\to x =\pm 1$
e) $A = 1$
$\to -\dfrac{6}{(x-2)(x+2)} = 1$
$\to x^2 - 4 = -6$
$\to x^2 = -2$ (vô nghiệm)
Vậy không có $x$ để $A = 1$
