a)

$\frac{x+2}{x-2}-$ $\frac{1-x}{x+2}=$ $\frac{3x^{2}+3+6}{x^{2}-4}$

⇔$\frac{x+2}{x-2}-$ $\frac{1-x}{x+2}=$ $\frac{3x^{2}+3+6}{x^{2}-x^{2}}$ 

⇔$\frac{x+2}{x-2}-$ $\frac{1-x}{x+2}=$ $\frac{3x^{2}+3+6}{(x+2)(x-2)}$ 

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

⇔$2x^{2}+x+6=$ $3x^{2}+3x+6$ 

⇔$2x^{2}+x=$ $3x^{2}+3x$ 

⇔$-x^{2}-2x=0$ 

⇔`x(x+2)=0`

->$\left[\begin{matrix}x=0 (t/m )\\ x=-2(không t/m)\end{matrix}\right.$

--> `x=0`

b)

$\frac{4}{x-3}+$ $\frac{3}{x+3}=$ $\frac{x-2}{x^{2}-4}$ 

⇔$\frac{4}{x-3}+$ $\frac{3}{x+3}=$ $\frac{x-2}{x^{2}-2^{2}}$ 

⇔$\frac{4}{x-3}+$ $\frac{3}{x+3}=$ $\frac{x-2}{(x-2)(x+2)}$ 

⇔$\frac{4}{x-3}+$ $\frac{3}{x+3}=$ $\frac{1}{x+2}$ 

⇔`4(x+3)(x+2) + 3(x-3)(x+2) = (x-3)(x+3)`

⇔$7x^{2}+17x+6=$ $x^{2}-9$ 

⇔$6x^{2}+17x+15=0$ 

⇔$6(x^{2}+$ $\frac{17}{6}x+$ $\frac{5}{2})=0$ 

⇔$6(x^{2}+$ $\frac{17}{6}x+$$\frac{289}{144}-$ $\frac{289}{144}+$$\frac{5}{2})=0$ 

⇔$6((x+\frac{17}{2})^{2}+$ $\frac{71}{144})=0$ 

⇔$6(x+\frac{17}{2})^{2}+$ $\frac{71}{24}=0$ 

⇔$(x+\frac{17}{2})^{2}=$ $\frac{-71}{144}$ 

Vô lí vì $(x+\frac{17}{2})^{2}$ $\geq0$ 

c)

$\frac{x}{x-5}+$ $\frac{x-3}{x+3}=$ $\frac{x^2+4}{x^2 -9}$ 

⇔$\frac{x}{x-5}+$ $\frac{x-3}{x+3}=$ $\frac{x^2+4}{x^2 -3^2}$ 

⇔$\frac{x}{x-5}+$ $\frac{x-3}{x+3}=$ $\frac{x^2+4}{(x-3)(x+3)}$ 

⇔`x(x+3)(x-3)+(x-3)^2(x-5)=` `(x^2+4)(x-5)`

⇔ `2x^3-11x^2+30x-45=x^3-5x^2+4x-20`

⇔ `x^3-6x^2+26x-25=0`

Giải thì ta ra được `1,24...`

-> `x=1,24`