a.Ta có :
$\left\{\begin{matrix} 2-x\neq 0 & \\ x^2-4\neq 0 & \\ 2+x\neq 0 & \\ 2x^2-x^3\neq 0 &\\ x^2-3x\neq 0 \end{matrix}\right.$ $⇔\left\{\begin{matrix} x\neq 2 & \\ x\neq ±2 & \\ x\neq -2 & \\ x\neq 0 ; x\neq 2 & \\ x\neq 3 \end{matrix}\right.$
Vậy ĐKXĐ của P là $x\neq ±2;x\neq 0;x\neq 3$
$b.P=\bigg(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\bigg) :\dfrac{x^2-3x}{2x^2-x^3} \\=\left [ \dfrac{(x+2)^2}{(2-x)(x+2)} +\dfrac{4x^2}{(2-x)(x+2)}+\dfrac{(2-x)^2}{(2-x)(x+2)}\right ]:\dfrac{x(x-3}{x^2(2-x)} \\=\dfrac{x^2+4x+4+4x^2-4+4x-x^2}{(2-x)(x+2)}.\dfrac{x(2-x)}{x-3} =\dfrac{4x(x+2)}{x+2}.\dfrac{x}{x-3} \\=\dfrac{4x^2}{x-3}$
$c.|x-5|=2$
⇔\(\left[ \begin{array}{l}x-5=2\\x-5=-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=7(TMĐKXĐ)\\x=3(KTMĐKXĐ)\end{array} \right.\)
$⇒P=\dfrac{4.7^2}{7-3}=49$
