\(\dfrac{x-2}{x+3}-\dfrac{x+1}{x-3}=\dfrac{30}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{x-2}{x+3}-\dfrac{x+1}{x-3}=\dfrac{30}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{30}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow x^2-3x-2x+6-\left(x^2+3x+x+3\right)=30\)
\(\Leftrightarrow x^2-3x-2x+6-x^2-3x-x-3=30\)
\(\Leftrightarrow-9x+3=30\)
\(\Leftrightarrow-9x-27=0\)
\(\Leftrightarrow x=-3\left(loại\right)\)
-Vậy S=∅
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