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

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

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

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

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

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

⇔$-x-1-1(x-3)=x(x+1)-(x-1)^2$

⇔$-x-1-x+3   =x^2+x-(x^2-2x+1)$

⇔$-2x+2        =x^2+x-x^2+2x-1  $

⇔$-2x+2        =                     3x-1  $

⇔$-2x-3x        =                       -2-1$

⇔$      -5x       =                           -3$

⇔$          x       =                      -3:-5$

⇔$          x       =             \frac{3}{5}$   

Vậy S={$\frac{3}{5}$}

F)$\frac{1}{x-2}$ - $\frac{6}{x+3}$ = $\frac{5}{6-x^2-x}$ 

⇔$\frac{1}{x-2}$ - $\frac{6}{x+3}$ = $\frac{5}{-x^2-x+6}$ 

⇔$\frac{1}{x-2}$ - $\frac{6}{x+3}$ = $\frac{5}{-x^2+2x-3x+6}$ 

⇔$\frac{1}{x-2}$ - $\frac{6}{x+3}$ = $\frac{5}{x(2-x)+3(2-x)}$ 

⇔$\frac{1}{x-2}$ - $\frac{6}{x+3}$ = $\frac{5}{(2-x)(x+3)}$ 

⇔$\frac{-1}{2-x}$ - $\frac{6}{x+3}$ = $\frac{5}{(2-x)(x+3)}$ 

⇔$\frac{-1(x+3)}{(2-x)(x+3)}$ - $\frac{6(2-x)}{(2-x)(x+3)}$ = $\frac{5}{(2-x)(x+3)}$ 

⇔$-1(x+3)-6(2-x)=5$

⇔$-x-3-12+6x=5$

⇔$-15+5x=5$

⇔$5x=5+15$

⇔$5x=20$

⇔$x=20:5$

⇔$x=4$

Vậy S={4}

G)$\frac{x+5}{x-5}$ - $\frac{x-5}{x+5}$=$\frac{20}{x^2-25}$ 

⇔$\frac{x+5}{x-5}$ - $\frac{x-5}{x+5}$=$\frac{20}{x^2-5^2}$ 

⇔$\frac{x+5}{x-5}$ - $\frac{x-5}{x+5}$=$\frac{20}{(x+5)(x-5)}$ 

⇔$\frac{(x+5)(x+5)}{(x+5)(x-5)}$ - $\frac{(x-5)(x-5)}{(x+5)(x-5)}$=$\frac{20}{(x+5)(x-5)}$ 

⇔$\frac{(x+5)^2}{(x+5)(x-5)}$ - $\frac{(x-5)^2}{(x+5)(x-5)}$=$\frac{20}{(x+5)(x-5)}$ 

⇔$\frac{x^2+10x+25}{(x+5)(x-5)}$ - $\frac{x^2-10x+25}{(x+5)(x-5)}$=$\frac{20}{(x+5)(x-5)}$ 

⇔$\frac{x^2+10x+25}{(x+5)(x-5)}$ - $\frac{x^2-10x+25}{(x+5)(x-5)}$=$\frac{20}{(x+5)(x-5)}$ 

⇔$x^2+10x+25-(x^2-10x+25)=20$

⇔$x^2+10x+25-x^2+10x-25=20$

⇔$20x=20$

⇔$x=20:20$

⇔$x=1$

Vậy S={1}

H)$\frac{1}{x-1}$ + $\frac{2}{x+1}$ = $\frac{x}{x^2-1}$ 

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

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

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

⇔$x+1+2x-2=x$ 

⇔$3x-1=x$ 

⇔$3x-x=1$ 

⇔$2x=1$ 

⇔$x=\frac{1}{2}$ 

Vậy S={ $\frac{1}{2}$ }

I)$5+\frac{76}{x^2-16} =\frac{2x-1}{x+4} - \frac{3x-1 }{4-x}$

⇔$5+\frac{76}{x^2-4^2} =\frac{2x-1}{x+4} - \frac{3x-1 }{4-x}$

⇔$\frac{5(x-4)(x+4)}{(x-4)(x+4)} +\frac{76}{(x-4)(x+4)} =\frac{(2x-1)(x-4)}{(x-4)(x+4)} - \frac{-(3x-1)(x+4) }{(x-4)(x+4)}$

⇔$5(x-4)(x+4)+76=(2x-1)(x-4)+(3x-1)(x+4)$

⇔$(5x-20)(x+4)+76=(2x-1)(x-4)+(3x-1)(x+4)$

⇔$5x^2+20x-20x-80+76=2x^2-8x-x+4+3x^2+12x-x-4$

⇔$5x^2-4=5x^2+2x$

⇔$5x^2-5x^2-2x=4$

⇔$-2x=4$

⇔$x=4:-2$

⇔$x=-2$

Vậy S={-2}

@minhthu6302007