Lời giải: 

a) ĐKXĐ: $\left \{ {{x-2≠0} \\{x+2≠0}\atop {x^2-4≠0}} \right.$  $⇔ \left \{ {{x≠2} \atop {x≠-2}} \right.$ 

Ta có: $\frac{1-6x}{x-2} + \frac{9x+4}{x+2} = \frac{x(3x-2)+1}{x^2-4}$ 
$⇔\frac{(1-6x)(x+2)}{x^2-4} + \frac{(9x+4)(x-2)}{x^2-4} = \frac{x(3x-2)+1}{x^2-4}$ 

$⇒x+2-6x^2-12x + 9x^2-18x+4x-8 = 3x^2-2x+1$

$⇔ -23x = 7$

$⇔ x = -\frac{7}{23}$ (TMĐKXĐ)
  Phương trình có tập nghiệm S = {$-\frac{7}{23}$}

b) ĐKXĐ: $\left \{ {{3-x≠0} \\{(x+2)(3-x)≠0} \atop {x+2≠0}} \right.$ $⇔\left \{ {{3-x≠0} \atop {x+2≠0}} \right.$ $⇔\left \{ {{x≠3} \atop {x≠-2}} \right.$ 

Ta có: $1 + \frac{x}{3-x} = \frac{5x}{(x+2)(3-x)} + \frac{2}{x+2}$ 

$⇔ \frac{(x+2)(3-x)}{(x+2)(3-x)} + \frac{x(x+2)}{(x+2)(3-x)} = \frac{5x}{(x+2)(3-x)} + \frac{2(3-x)}{(x+2)(3-x)}$ 

$⇒ 3x-x^2+6-2x + x^2+2x = 5x + 6-2x$

$⇔ 3x + 6 = 3x + 6$

$⇔ 3x - 3x = 6 - 6$

$⇔ 0x = 0$
    Phương trình có vô số nghiệm

c) ĐKXĐ: $\left \{ {{x-1≠0} \\{x^2+x+1≠0}\atop {x^3-1≠0}} \right.$  $⇔ x≠1$

Ta có: $\frac{2}{x-1} + \frac{2x+3}{x^2+x+1} = \frac{(2x+1)(2x-1)}{x^3-1}$ 

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

$⇒ 2x^2+2x+2 + 2x^2-2x+3x-3 = 4x^2-2x+2x-1$

$⇔4x^2+3x-1 = 4x^2-1$
 
$⇔ 3x = 0$  

$⇔ x = 0$ (TMĐKXĐ)

      Phương trình có tập nghiệm S = {0}

d) ĐKXĐ: $\left \{ {{(4x+3)(x-5)≠0} \\{4x+3≠0}\atop {x-5≠0}} \right.$ $⇔\left \{ {{4x≠-3} \atop {x≠5}} \right.$ $⇔\left \{ {{x≠\frac{-3}{4}} \atop {x≠5}} \right.$ 

Ta có: $\frac{x^3-(x-1)^3}{(4x+3)(x-5)} = \frac{7x-1}{4x+3} - \frac{x}{x-5}$ 

$⇔\frac{x^3-(x-1)^3}{(4x+3)(x-5)} = \frac{(7x-1)(x-5)}{(4x+3)(x-5)} - \frac{x(4x+3)}{(4x+3)(x-5)}$ 

$⇒ x^3-(x-1)^3 = 7x^2-35x-x+5 - 4x^2-3x$

$⇔ x^3 - x^3 + 3x^2 - 3x + 1 = 3x^2 - 39x + 5$

$⇔ 36x = 4$

$⇔ x = \frac{1}{9}$ (TMĐKXĐ)
    Phương trình có tập nghiệm S = {$\frac{1}{9}$} 

`a)` `frac{1-6x}{x-2}+frac{9x+4}{x+2}=frac{x(3x-2)+1}{x^2-4}`   ĐKXĐ: `x\ne±2`

`<=>frac{1-6x}{x-2}+frac{9x+4}{x+2}-frac{3x^2-2x+1}{x^2-4}=0`

`<=>frac(x+2)(1-6x)+(x-2)(9x+4)-(3x^2-2x+1)(x-2)(x+2)=0`

`=>(x+2)(1-6x)+(x-2)(9x+4)-(3x^2-2x+1)(x-2)(x+2)=0`

`<=>x-6x^2+2-12x+9x^2+4x-18x-8-3x^2+2x-1=0`

`<=>-23x-7=0`

`<=>-23x=7`

`<=>x=-7/23`  (TMĐK)

Vậy `S={-7/23}`

`b)` `1+frac{x}{3-x}=frac{5x}{(x+2)(3-x)}+frac{2}{x+2}`   ĐKXĐ: `x\ne-2;x\ne3`

`<=>frac{(3-x)(x+2)+x(x+2)}{(3-x)(x+2)}=frac{5x}{(x+2)(3-x)}+frac{2(3-x)}{(x+2)(3-x)}`

`=>(3-x)(x+2)+x(x+2)=5x+2(3-x)`

`<=>3x+6-x^2-2x+x^2+2x=5x+6-2x`

`<=>3x+6=3x+6`

`<=>3x+6-3x-6=0`

`<=>0x=0`

Vậy phương trình trên có vô số nghiệm khi `x\ne-2;x\ne3`

`c)` `frac{2}{x-1}+frac{2x+3}{x^2+x+1}=frac{(2x-1)(2x+1)}{x^3-1}`   ĐKXĐ: `x\ne1`

`<=>frac{2(x^2+x+1)+(x-1)(2x+3)}{(x-1)(x^2+x+1)}=frac{(2x-1)(2x+1)}{(x-1)(x^2+x+1)}`

`=>2x^2+2x+2+(2x^2+3x-2x-3)=4x^2-1`

`<=>2x^2+2x+2+2x^2+x-3=4x^2-1`

`<=>4x^2+3x-1=4x^2-1`

`<=>4x^2+3x-1-4x^2+1=0`

`<=>3x=0`

`<=>x=0`  (TMĐK)

Vậy phương trình có nghiệm `x=0`

`d)` `frac{x^3-(x-1)^3}{(4x+3)(x-5)}=frac{7x-1}{4x+3}-frac{x}{x-5}`   ĐKXĐ: `x\ne-3/4;x\ne5`

`<=>frac{x^3-(x-1)^3}{(4x+3)(x-5)}=frac{(7x-1)(x-5)-x(4x+3)}{(x-5)(4x+3)}`

`=>x^3-(x-1)^3=(7x-1)(x-5)-x(4x+3)`

`<=>x^3-(x^3-3x^2+3x-1)=7x^2-x-35x+5-4x^2-3x`

`<=>x^3-x^3+3x^2-3x+1=3x^2-39x+5`

`<=>3x^2-3x+1=3x^2-39x+5`

`<=>3x^2-3x+1-3x^2+39x-5=0`

`<=>36x-4=0`

`<=>36x=4`

`<=>x=1/9`  `(TMĐK)`

Vậy `S={1/9}`