Question

giải pt

a. \(\dfrac{1}{x-4}+\dfrac{1}{x+4}=\dfrac{1}{3}\)

Answer 1

điều kiện : \(x\ne\pm4\)

ta có : \(\dfrac{1}{x-4}+\dfrac{1}{x+4}=\dfrac{1}{3}\) \(\Leftrightarrow\dfrac{x+4+x-4}{\left(x-4\right)\left(x+4\right)}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{2x}{x^2-16}=\dfrac{1}{3}\Leftrightarrow2x=\dfrac{1}{3}\left(x^2-16\right)\)

\(\Leftrightarrow6x=x^2-16\Leftrightarrow x^2-6x-16=0\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\) vậy \(x=8;x=-2\)