a, 9(x+2)-x²(x+2) =0
⇔(x+2)(9-x²)=0
⇔(x+2)(3-x)(3+x)=0
⇒\(\left[ \begin{array}{l}x+2=0\\3-x=0\\3+x=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=-2\\x=3\\x=-3\end{array} \right.\)
vậy S={-2 ; -3 ; 3}
b, |x+3|-2=3x
⇔|x+3|=3x+2
⇔\(\left[ \begin{array}{l}x+3=3x+2\\x+3=-3x-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{1}{2} \\x=\frac{-5}{3} \end{array} \right.\)
vậy S={$\frac{1}{2}$ ; $\frac{-5}{3}$}
c, $\frac{x}{x-4}$+ $\frac{x-3}{x+4}$=$\frac{x²+12}{x²-16}$
⇔x²+4x+x²-7x+12=x²+12
⇔x²-3x=0
⇔x(x-3)=0
⇒\(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
vậy S={0;3}
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