\(\Leftrightarrow\dfrac{3}{x-4}+\dfrac{4}{x-4}-\dfrac{3x-4}{x^2-16}=0\left(dkxd:x\ne\pm4\right)\)
\(\Leftrightarrow3\left(x+4\right)+4\left(x+4\right)-3x+4=0\)
\(\Leftrightarrow3x+12+4x+16-3x+4=0\)
\(\Leftrightarrow4x+32=0\)
\(\Leftrightarrow x=-8\left(tm\right)\)
Vậy \(S=\left\{-8\right\}\)
\(ĐK:\left\{{}\begin{matrix}x\ne4\\x\ne-4\end{matrix}\right.\\ \Leftrightarrow\dfrac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\\ \Leftrightarrow3\left(x+4\right)-4\left(x-4\right)=3x-4\\ \Leftrightarrow3x+12-4x+16-3x+4=0\\ \Leftrightarrow-4x+32=0\\ \Leftrightarrow-4x=-32\\ \Leftrightarrow x=8\left(t/m\right)\)
\(\dfrac{3}{x-4}-\dfrac{4}{4-x}=\dfrac{3x-4}{x^2-16}\left(Dk:x\ne\pm4\right)\)
\(\Leftrightarrow\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow\dfrac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow3x+12-4x+16=3x-4\)
\(\Leftrightarrow3x-4x-3x=-4-16-12\)
\(\Leftrightarrow-4x=-32\)
\(\Leftrightarrow x=8\left(Thoaman\right)\)
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