\(đk:\left\{{}\begin{matrix}x-4\ne0\\x+2\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ne4\\x\ne-2\end{matrix}\right.\\ \dfrac{2}{x-4}-\dfrac{5}{x+2}=\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}\\ \Leftrightarrow\dfrac{2.\left(x+2\right)-5.\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}=\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}\\ \Leftrightarrow2x+4-5x+20=3x-5\\ \Leftrightarrow-3x+24=3x-5\\ \Leftrightarrow-6x=-29\\ \Leftrightarrow x=\dfrac{29}{6}\left(t/m\right)\)
\(\dfrac{2}{x-4}-\dfrac{5}{x+2}=\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}\left(dkxd:x\ne4;x\ne-2\right)\)
\(\Leftrightarrow\dfrac{2}{x-4}-\dfrac{5}{x+2}-\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{2\left(x+2\right)-5\left(x-4\right)-\left(3x-5\right)}{\left(x-4\right)\left(x+2\right)}=0\)
\(\Leftrightarrow2x+4-5x+20-3x+5=0\)
\(\Leftrightarrow-6x=-29\)
\(\Leftrightarrow x=\dfrac{29}{6}\left(tmdk\right)\)
Vậy \(S=\left\{\dfrac{29}{6}\right\}\)
Giải:
ĐKXĐ:x≠4;x≠-2
MSC:(x-4)(x+2)
<=> 2(x+2)/(x-4)(x+2)-5(x-4)/(x-4)(x+2)=3x-5/(x-4)(x+2)
=>2x+4-5x+20=3x-5
<=> 2x-5x-3x=-5-4-20
<=> -6x=-29
<=> x=29/6
Vậy:S={29/6}
\(\dfrac{2}{x-4}-\dfrac{5}{x+2}=\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x-4\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne-2\end{matrix}\right.\)
\(\dfrac{2}{x-4}-\dfrac{5}{x+2}=\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}\)
⇔\(\dfrac{2\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}-\dfrac{5\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}=\dfrac{3x-5}{\left(x-4\right)\left(x+2\right)}\)
⇒\(2x+4-5x+20=3x-5\)
⇔\(2x-5x-3x=-5-4-20\)
⇔\(-6x=-29\)
⇔\(x=-\dfrac{29}{6}\)
Vậy phương trình có nghiệm là: \(x=-\dfrac{29}{6}\)