Tính: 2x^2/x^2-1 - 6/3-x - 2x-6/x^3-3x^2-x+3

Đáp án:

$\begin{array}{l}
\frac{{2{x^2}}}{{{x^2} - 1}} - \frac{6}{{3 - x}} - \frac{{2x - 6}}{{{x^3} - 3{x^2} - x + 3}}\\
 = \frac{{2{x^2}}}{{{x^2} - 1}} + \frac{6}{{x - 3}} - \frac{{2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {{x^2} - 1} \right)}}\\
 = \frac{{2{x^2}}}{{{x^2} - 1}} + \frac{6}{{x - 3}} - \frac{2}{{{x^2} - 1}}\\
 = \frac{{2{x^2} - 2}}{{{x^2} - 1}} + \frac{6}{{x - 3}}\\
 = 2 + \frac{6}{{x - 3}}\\
 = \frac{{2x - 6 + 6}}{{x - 3}}\\
 = \frac{{2x}}{{x - 3}}
\end{array}$

$\frac{2x^2}{x^2-1}-$ $\frac{6}{3-x}-$ $\frac{2x-6}{x^3-3x^2-x+3}$

$⇔\frac{2x^2}{x^2-1 }+$ $\frac{6}{x-3}-$ $\frac{2(x-3)}{(x-3)(x^2-1)}$

$⇔\frac{2x^2}{x^2-1}+$ $\frac{6}{x-3}-$ $\frac{x}{x^2-1}$

$⇔\frac{2x^2-2}{x^2-1}+$ $\frac{6}{x-3}$

$⇔2+\frac{6}{x-3}$

$⇔\frac{2x-6+6}{x-3}$

$⇔\frac{2x}{x-3}$