a)2x²-x=3-6x
<=>2²-x+6x-3=0
<=>2x²+5x-3=0
<=>2x²+6x-x-3=0
<=>2x(x+3)-(x+3)=0
<=>(x-3)(2x-1)=0
TH1:x-3=0
<=>x=3
TH2:2x-1=0
<=>2x=1
<=>x=$\frac{1}{2}$
Vậy S ={
$\frac{1}{2}$ ;3}
b)$\frac{x-2}{x+2}$ + $\frac{3}{2-x}$ = $\frac{2(x-11)}{x^2-4}$ ĐKXĐ:x khác +-2
<=>$\frac{(x-2)(x-2)}{(x-2)(x+2)}$ - $\frac{3}{x-2}$ = $\frac{2x-22}{(x-2)(x+2)}$
<=>$\frac{(x-2)^2}{(x-2)(x+2)}$ - $\frac{3(x+2)}{(x-2)(x+2)}$ = $\frac{2x-22}{(x-2)(x+2)}$
<=>(x-2)²-3(x+2)=2x-22
<=>x²-4x+4-3x-6=2x-22
<=>x²-4x+4-3x-6-2x+22=0
<=>x²-9x+20=0
<=>x²-5x-4x+20=0
<=>x(x-5)-4(x-5)=0
<=>(x-5)(x-4)=0
TH1:x-5=0
<=>x=5(tmđk)
TH2:x-4=0
<=>x-4(tmđk)
Vậy S={4;5}
