a) \(\dfrac{3}{x}-5=-\dfrac{4}{x}+2\)
\(\dfrac{3}{x}+\dfrac{4}{x}=7\)
\(\dfrac{7}{x}=7\Rightarrow x=1\)
b) \(\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=4\) hoặc \(x=-4\)
Ta có : \(\dfrac{3}{x}+\dfrac{4}{x}=2+5=7\)
x =1