Giải các phương trình sau: a) 3(x – 5) + 2(x + 7) = x + 11.

d) $\frac{2\text{x}-3}{x+2}+\frac{\text{x}-5}{3-x}-\frac{10}{x^2-x-6}=2$

ĐKXĐ: $$\begin{gathered} \left\{\begin{array}{l}x+2\ne 0\\ 3-x\ne 0\\ x^2-x-6\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x+2\ne 0\\ 3-x\ne 0\\ (x+2)(x-3)\ne 0\end{array}\right. \\ \iff \left\{\begin{array}{l}x\ne -2\\ x\ne 3\end{array}\right. \end{gathered}$$

Khi đó phương trình đã cho trở thành:

$$\begin{gathered} \frac{2\text{x}-3}{x+2}-\frac{\text{x}-5}{x-3}-\frac{10}{(x+2)(x-3)}=2 \\ \iff \frac{\left(2x-3\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}+\frac{\left(\text{x}-5\right)\left(-1\right)\left(x+2\right)}{\left(3-x\right)\left(-1\right)\left(x+2\right)}-\frac{10}{\left(x+2\right)\left(x-3\right)}=2.\frac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)} \\ \iff \frac{\left(2x-3\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}+\frac{\left(-x-2\right)\left(x-5\right)}{\left(x-3\right)\left(x+2\right)}-\frac{10}{\left(x+2\right)\left(x-3\right)}=2.\frac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)} \end{gathered}$$

=> (x – 3)(2x – 3) + (x – 5)(–2 – x) – 10 = 2(x + 2)(x – 3)

Û 2x² – 9x + 9 – x² + 10 + 3x – 10 = 2(x² – x – 6)

Û x² – 6x + 9 = 2x² – 2x – 12

Û 2x² – x² – 2x + 6x – 12 – 9 = 0

Û x² + 4x – 21 = 0

Û x² + 7x – 3x – 21 = 0

Û (x² + 7x) – (3x + 21) = 0

Û x(x + 7) – 3(x + 7) = 0

Û (x + 7)(x – 3) = 0

$$\begin{gathered} \iff \left[\begin{array}{l}-x-7=0\\ x-3=0\end{array}\right. \\ \iff \left[\begin{array}{l}x=-7\\ x=3\end{array}\right. \end{gathered}$$