Giải các phương trình sau: a) 2x – 6 = x + 1.

d) $\frac{3}{x-2}-\frac{5}{x+2}=\frac{8\text{x}-14}{x^2-4}$

ĐKXĐ: $$\begin{gathered} \left\{\begin{array}{l}x-2\ne 0\\ x+2\ne 0\\ x^2-4\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x-2\ne 0\\ x+2\ne 0\\ (x+2)(x-2)\ne 0\end{array}\right. \\ \iff \left\{\begin{array}{l}x\ne 2\\ x\ne -2\end{array}\right.\iff x\ne \pm 2 \end{gathered}$$

Khi đó phương trình đã cho tương đương với:

$$\begin{gathered} \frac{3}{x-2}-\frac{5}{x+2}=\frac{8\text{x}-14}{\left(x-2\right)\left(x+2\right)} \\ \iff \frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8\text{x}-14}{\left(x-2\right)\left(x+2\right)} \end{gathered}$$

Þ 3(x + 2) – 5(x – 2) = 8x – 14

Û 3x + 6 – 5x + 10 = 8x – 14

Û 3x – 5x – 8x = – 14 – 6 – 10

Û – 10x = – 30

Û x = 3 (thỏa mãn)