\(\dfrac{9}{x+4}=\dfrac{3}{x-2}\) (ĐK: \(x\ne2;-4\))
\(\Rightarrow9\left(x-2\right)=3\left(x+4\right)\)
\(\Leftrightarrow9x-18=3x+12\)
\(\Leftrightarrow9x-3x=12+18\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\) (TMĐK)
Vậy \(S=\left\{5\right\}\).
\(\dfrac{9}{x+4}=\dfrac{3}{x-2}\text{ĐKXĐ:}x\ne-4;2\)
\(\Leftrightarrow\dfrac{9\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}=\dfrac{3\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}MTC:\left(x+4\right)\left(x-2\right)\)
\(\Rightarrow9x-18=3x+12\)
\(\Leftrightarrow9x-18-3x-12=0\)
\(\Leftrightarrow6x-30=0\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{5\right\}\)
