7. x 2 − x − 12 = 0

$\frac{1}{x-1}+\frac{2}{x+1}=\frac{x}{x^2-1}$ (ĐKXĐ: $x\ne \pm 1$)

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

$S=\left\{\frac{1}{2}\right\}$

$\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}$ (ĐKXĐ: $x\ne \pm 5$)

$\begin{array}{l}\iff \frac{{\left(x+5\right)}^2-{\left(x-5\right)}^2}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \iff \frac{x^2+10x+25-x^2+10x-25}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \iff \frac{20x}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \Rightarrow 20x=20\iff x=1\left(TM\right)\\ S=\left\{1\right\}\end{array}$

$\frac{2}{x-1}+\frac{4}{x+3}=\frac{3x+11}{x^2+2x-3}$ (ĐKXĐ: $x\ne 1,x\ne -3$)

$\begin{array}{l}\iff \frac{2\left(x+3\right)+4\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{3x+11}{\left(x-1\right)\left(x+3\right)}\\ \iff \frac{2x+6+4x-4}{\left(x-1\right)\left(x+3\right)}=\frac{3x+11}{\left(x-1\right)\left(x+3\right)}\\ \Rightarrow 6x+2=3x+11\\ \iff 3x=9\\ \iff x=3\left(TM\right)\end{array}$

$S=\left\{3\right\}$

Đặt $2x^2+3x=A$ ta có:

$\begin{array}{l}{\left(A-1\right)}^2-5.\left(A+3\right)+24=0\\ \iff A^2-7A+10=0\\ \iff \left(A-5\right)\left(A-2\right)=0\\ \Rightarrow \left[\begin{array}{l}A-5=0\\ A-2=0\end{array}\right.\iff \left[\begin{array}{l}A=5\\ A=2\end{array}\right.\end{array}$

TH1: Với $A=5\Rightarrow 2x^2+3x=5$

$\begin{array}{l}\iff 2x^2+3x-5=0\\ \iff \left(x-1\right)\left(2x+5\right)=0\\ \iff \left[\begin{array}{l}x=1\\ x=\frac{-5}{2}\end{array}\right.\end{array}$

TH2: Với $A=2\Rightarrow 2x^2+3x=2$

$\begin{array}{l}\iff 2x^2+3x-2=0\\ \iff \left(2x-1\right)\left(x+2\right)=0\\ \Rightarrow \left[\begin{array}{l}x=\frac{1}{2}\\ x=-2\end{array}\right.\end{array}$

$S=\left\{\frac{-5}{2};-2;\frac{1}{2};1\right\}$

$x^3-3x^2+4=0$

$$\begin{gathered} \begin{array}{l}\iff \left(x+1\right)\left(x^2-4x+4\right)=0\\ \iff \left(x+1\right){\left(x-2\right)}^2=0\\ \iff \left[\begin{array}{l}x+1=0\\ {\left(x-2\right)}^2=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=-1\\ x=2\end{array}\right.\end{array} \\ S=\left\{-1;2\right\} \end{gathered}$$