Ta có: \(\dfrac{x+5}{x-5}-\dfrac{1}{x}=\dfrac{5}{x\left(x-5\right)}\)
Suy ra: \(x^2+5x-x+5=5\)
\(\Leftrightarrow x\left(x+4\right)=0\)
hay x=-4
\(Đk:\)\(x\)\(\ne\)\(0\) và x\(\ne\)\(5\)
\(\Leftrightarrow\)\(\dfrac{x(x+5)-(x-5)}{x(x-5)}=\dfrac{5}{x(x-5)}\)
\(\Leftrightarrow\)\(\dfrac{x^2+5x-x+5}{x(x-5)}-\dfrac{5}{x(x-5)}=0\)
\(\Leftrightarrow\)\(\dfrac{x^2+4x}{x(x-5)}=0\)
\(\Rightarrow\)\(x^2+4x=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=0\\ x+4=0 \end{array} \right.\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=0(loại)\\ x=-4(thỏa mãn) \end{array} \right.\)
