Answer: x ≠ 5; x≠ 0
[tex]\frac{4}{x-5}+\frac{1}{x}=\frac{x-1}{x-5}\\\\<=> \frac{4x+x-5}{x(x-5)}=\frac{x(x-1)}{x(x-5)} \\<=> 4x+x-5=x(x-1)\\<=> 5x-5 = x^{2}-x\\<=> x^{2} -6x+5=0\\<=>(x-5)(x-1)=0\\[/tex]
because x≠ 5 => x - 5 ≠ 0
=> x - 1 = 0
⇔ x = 1
Step-by-step explanation:
