a, $\frac{x-1}{x+2}-$ $\frac{x}{x-2}=$ $\frac{5x-2}{4-x^2}$
⇔$\frac{(x-1)(x-2)-x(x+2)}{(x-2)(x+2)}=$ $\frac{2-5x}{x^2-4}$
⇔$\frac{(x-1)(x-2)-x(x+2)-2+5x}{x^2-4}=0$
⇔$(x-1)(x-2)-x(x+2)-2+5x=0$
⇔$x^2-3x+2-x^2-2x-2+5x=0$
⇔$0x-0=0$
⇔$0x=0$
Vậy S=R
b, $\frac{x-3}{x-2}-$ $\frac{x-2}{x-4}=$ $3.\frac{1}{5}$
⇔$\frac{(x-3)(x-4)-(x-2)(x-2)}{(x-2)(x-4)}=$ $\frac{16}{5}$
⇔$\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=$ $\frac{3}{5}$
⇔$\frac{-3x+8}{x^2-6x+8}=$ $\frac{3}{5}$
⇔$-15x+40=3x^2-18x+24$
⇔$-15x+40-3x^2+18x-24=0$
⇔$3x-3x^2+16=0$
⇔$x=$$\frac{1}{2}-$ $\frac{\sqrt[]{201}}{6}$ hoặc $x=$$\frac{\sqrt[]{201}}{6}+$ $\frac{1}{2}(tm)$
c, $\frac{2x+1}{x-1}=$ $\frac{5(x-1)}{x+1}$
⇔$(2x+1)(x+1)=5(x-1)(x-1)$
⇔$2x^2+2x+x+1=5(x-1)^2$
⇔$2x^2+2x+x+1-5x^2+10x-5=0$
⇔$-3x^2+13x-4=0$
⇔$x=$±$\frac{\sqrt[]{313}-13}{18}(tm)$
