Đáp án:

 $a)  x=0\\
b) {\left[\begin{aligned}x=0\\x=5\end{aligned}\right.}\\
c) S=\varnothing \\
d) x=\frac{-3}{2}\\
e) 
 \forall x\in \mathbb{R}$

Giải thích các bước giải:

$a) \frac{x+6}{x-5}+\frac{x-5}{x+6}=\frac{2x^2+23x+61}{x^2+x-30}\\
ĐK: {\left\{\begin{aligned}x-5\neq 0\\x+6\neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x\neq 5\\x\neq -6\end{aligned}\right.}\\
\Rightarrow \frac{(x+6)^2}{(x+6)(x-5)}+\frac{(x-5)^2}{(x-5)(x+6)}=\frac{2x^2+23x+61}{(x-5)(x+6)}\\
\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\\
\Leftrightarrow 2x^2+2x+61-2x^2-23x-61=0\\
\Leftrightarrow -21x=0\\
\Leftrightarrow x=0\\
b) \frac{6}{x-5}+\frac{x+2}{x-8}=\frac{18}{(x-5)(8-x)}-1\\
ĐK: {\left\{\begin{aligned}x+2\neq 0\\x-8 \neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x\neq -2\\x \neq 8\end{aligned}\right.}\\
\Rightarrow \frac{6(x-8)}{(x-5)(x-8)}+\frac{(x-5)(x+2)}{(x-5)(x-8)}=\frac{-18}{(x-5)(x-8)}-\frac{(x-5)(x-8)}{(x-5)(x-8)}\\
\Leftrightarrow 6x-48+x^2+2x-5x-10=-18-(x^2-8x-5x+40)\\
\Leftrightarrow x^2+3x-58+18+x^2-8x-5x+40=0\\
\Leftrightarrow 2x^2-10x=0\\
\Leftrightarrow 2x(x-5)=0\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x-5=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x=5\end{aligned}\right.}\\
c) \frac{x-4}{x-1}+\frac{x+4}{x+1}=2\\
ĐK: {\left\{\begin{aligned}x-1\neq 0\\x+1\neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x\neq 1\\x\neq -1\end{aligned}\right.}\\
\Rightarrow \frac{(x+1)(x-4)}{(x+1)(x-1)}+\frac{(x-1)(x+4)}{(x-1)(x+1)}=\frac{2(x^2-1)}{x^2-1}\\
\Leftrightarrow x^2-4x+x-4+x^2+4x-x-4=2x^2-2\\
\Leftrightarrow 2x^2-8-2x^2+2=0$
$\Leftrightarrow -10=0$ (vô lý)
$\Rightarrow S=\varnothing \\
d) \frac{3}{x+1}-\frac{1}{x-2}=\frac{9}{(x+1)(2-x)}\\
ĐK: {\left\{\begin{aligned}x-2\neq 0\\x+1\neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x\neq 2\\x\neq -1\end{aligned}\right.}\\
\Rightarrow \frac{3(x-2)}{(x+1)(x-2)}-\frac{x+1}{(x+1)(x-2)}=\frac{-9}{(x+1)(x-2)}\\
\Leftrightarrow 3x-6-x=-9\\
\Leftrightarrow 2x=-9+6=-3\\
\Leftrightarrow x=\frac{-3}{2}\\
e) \frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x}{9-x^2}\\
ĐK: {\left\{\begin{aligned}x-3\neq 0\\x+3\neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x\neq 3\\x\neq -3\end{aligned}\right.}\\
\Rightarrow \frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{-(7x^2-3x)}{x^2-9}\\
\Leftrightarrow \frac{(x-3)(x^2-x)}{x^-9}-\frac{x^2(x+3)}{x^2-9}=\frac{-(7x^2-3x)}{x^2-9}\\
\Leftrightarrow x^3-x^2-3x^2+3x-x^3-3x^2=-(7x^2-3x)\\
 \Leftrightarrow x^3-x^2-3x^2+3x-x^3-3x^2+7x^2-3x=0\\
 \Leftrightarrow 0x=0\\
 \forall x\in \mathbb{R}$