\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)
suy ra: \(3\left(x-3\right)=5\left(x+2\right)\\ < =>3x-9=5x+10\\ < =>3x-5x=10+9\\ < =>-2x=19\\ < =>x=-\dfrac{19}{2}\left(tm\right)\)
\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\)ĐKXĐ \(\left\{{}\begin{matrix}x+2\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\)
`<=> 3(x-3) =5 (x+2)`
`<=> 3x-9 = 5x+10`
`<=>3x -5x=10+9`
`<=> -2x=19`
`<=>x=-19/2`
\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\text{ĐKXĐ}:x\ne-2;3\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}MTC:\left(x+2\right)\left(x-3\right)\)
\(\Rightarrow3x-9=5x+10\)
\(\Leftrightarrow3x-9-5x-10=0\)
\(\Leftrightarrow-2x-19=0\)
\(\Leftrightarrow-2x=19\)
\(\Leftrightarrow x=\dfrac{-19}{2}\left(\text{nhận}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-19}{2}\right\}\)