`x/[x-3]-[2x+15]/[x^2-9]=5/[x+3]` `ĐK: x \ne +-3`
`<=>[x(x+3)-2x-15]/[(x-3)(x+3)]=[5(x-3)]/[(x-3)(x+3)]`
`=>x^2+3x-2x-15=5x-15`
`<=>x^2-4x=0`
`<=>x(x-4)=0`
`<=>` $\left[\begin{matrix} x=0\\ x-4=0\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=0\\ x=4\end{matrix}\right.$ (t/m)
Vậy `S={0;4}`
ĐKXĐ : x ≠ 3; x ≠ - 3
\(\Rightarrow x\left(x+3\right)-2x-15=5\left(x-3\right)\)
\(\Leftrightarrow x^2+3x-2x-15=5x-15\)
\(\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\) (N)
\(\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x+15}{\left(x-3\right)\left(x+3\right)}=\dfrac{5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x^2+3x-2x+15-5x-15=0\)
\(\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
