Rút gọn P = [$\frac{3+x}{3-x}$ - $\frac{3-x}{3+x}$ + $\frac{4x²}{x²-9}$] : [$\frac{2x+1}{x+3}$ - 1]

Đáp án:

$P=\dfrac{4x}{x-2}.$

Giải thích các bước giải:

$P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}+\dfrac{4x^2}{x^2-9}\right): \left(\dfrac{2x+1}{x+3}-1\right)\\ \text{ĐKXĐ: }\left\{\begin{array}{l} 3-x \ne 0 \\ 3+x \ne 0 \\ x^2-9 \ne 0 \\ \dfrac{2x+1}{x+3}-1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne 3 \\ x \ne -3 \\ x^2 \ne 9 \\ \dfrac{2x+1-x-3}{x+3} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne \pm 3 \\ \dfrac{x-2}{x+3} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne \pm 3 \\ x-2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne \pm 3 \\ x \ne 2\end{array} \right.\\ P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}+\dfrac{4x^2}{x^2-9}\right): \left(\dfrac{2x+1}{x+3}-1\right)\\ =\left(\dfrac{-3-x}{x-3}-\dfrac{3-x}{x+3}+\dfrac{4x^2}{(x-3)(x+3)}\right): \dfrac{x-2}{x+3}\\ =\left(\dfrac{(-3-x)(x+3)}{(x-3)(x+3)}-\dfrac{(3-x)(x-3)}{(x+3)(x-3)}+\dfrac{4x^2}{(x-3)(x+3)}\right). \dfrac{x+3}{x-2}\\ =\dfrac{-(x+3)^2+(x-3)^2+4x^2}{(x-3)(x+3)}. \dfrac{x+3}{x-2}\\ =\dfrac{-(x^2+6x+9)+x^2-6x+9+4x^2}{(x-3)(x+3)}. \dfrac{x+3}{x-2}\\ =\dfrac{4 x^2 - 12 x}{(x-3)(x+3)}. \dfrac{x+3}{x-2}\\ =\dfrac{4 x(x - 3)}{(x-3)(x+3)}. \dfrac{x+3}{x-2}\\ =\dfrac{4x}{x-2}.$