$\frac{x}{2(x-3)}$ + $\frac{x}{2(x+1)}$ = $\frac{2x}{(x+1)(x+3)}$ ĐKXĐ:x$\neq$ -1;x$\neq$ ±3
⇔$\frac{x(x+3)(x+1)}{2(x-3)(x+3)(x+1)}$ + $\frac{x(x+3)(x-3)}{2(x+1)(x+3)(x-3)}$ = $\frac{4x(x-3)}{2(x+1)(x+3)(x-3)}$
=>x(x+3)(x+1)+x(x+3)(x-3)=4x(x-3)
⇔x(x²+x+3x+3)+x(x²-9)=4x²-12x
⇔x³+x²+3x²+3x+x³-9x-4x²+12x=0
⇔2x³+6x=0
⇔2x(x²+3)=0
TH1:2x=0
⇔x=0(t/m)
TH2:x²+3=0
⇔x²=-3 (vô lí)
=> x∈ rỗng
Vậy S∈{0}
