a, $\frac{1-x}{x+1}+3=$ $\frac{2x+3}{x+1}$
=> $\frac{1-x+3(x+1)-(2x+3)}{x+1}=0$
=>$1-x+3x+3-2x-3=0$
=>$0x=-1$(loại)
Vậy S=∅
b, $\frac{(x+2)^2}{2x-3}-1=$ $\frac{x^2+10}{2x-3}$
=> $\frac{(x+2)^2-(2x-3)-(x^2+10)}{2x-3}=0$
=>$x^2+4x+4-2x+3-x^2-10=0$
=>$2x-3=0$
=>$x=3/2$
Vậy S={3/2}
c, $\frac{x-1}{x-2}-$ $\frac{5}{x+2}=$ $\frac{12}{x^2-4}+1$
=>$\frac{(x-1)(x+2)-5(x-2)-12-(x^2-4)}{(x-2)(x+2)}=0$
=>$x^2+x-2-5x+10-12-x^2+4=0$
=>$-4x=0$(loại)
Vậy S=∅
d, $\frac{x+1}{x-2}-$ $\frac{2}{x^2-2x}=$ $\frac{1}{x}$ =>$\frac{(x+1)x-2-(x-2)}{x(x-2)}=0$=>$x^2+x-2-x+2=0$=>$x^2=0$=>$x=0$
Vậy S={0}
e, $\frac{3x+2}{3x-2}-$ $\frac{6}{2+3x}=$ $\frac{9x^2}{9x^2-4}$
=> $\frac{(3x+2)(3x+2)-6(3x-2)-9x^2}{(3x-2)(3x+2)}=0$
=>$(3x+2)^2-18x+12-9x^2=0$
=>$9x^2+12x+4-18x+12-9x^2=0$
=>$-6x=-16$=>$x=8/3$
Vậy S={8/3}
f, $\frac{12}{1-9x^2}=$ $\frac{1-3x}{1+3x}-$ $\frac{1+3x}{1-3x}$ =>$\frac{12-(1-3x)(1-3x)+(1+3x)(1+3x)}{1-9x^2}=0$=>$12-(1-3x)^2+(1+3x)^2=0$=>$12+6x=0$=>$x=-2$
Vậy S+{-2}
