Giúp mình 2 câu nayc với ạ

a)$\frac{201-x}{99}$+$\frac{203-x}{97}$+$\frac{205-x}{95}$+3=0
⇔($\frac{201-x}{99}$+1)+($\frac{203-x}{97}$+1)+($\frac{205-x}{95}$+1)=0
⇔$\frac{201-x+99}{99}$+$\frac{203-x+97}{97}$+$\frac{205-x+95}{95}$=0
⇔$\frac{300-x}{99}$+$\frac{300-x}{97}$+$\frac{300-x}{95}$=0
⇔(300-x)($\frac{1}{99}$+$\frac{1}{97}$+$\frac{1}{95}$)=0
⇔300-x=0
⇔x=300
Vậy S={300}
b)  $\frac{x²+x+4}{2}$ +$\frac{x²+x+7}{3}$=$\frac{x²+x+13}{5}$+$\frac{x²+x+16}{6}$ 
⇔15x²+15x+60+10x²+10x+70=6x²+6x+78+5x²+5x+80
⇔14x²+14x-28=0
⇔14(x²+x-2)=0
⇔14[(x²-x)+(2x-2)]=0
⇔14[x(x-1)+2(x-1)]=0
⇔14(x-1)(x+2)=0

⇔\(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\) 

Vậy S={1;-2}.