\(\frac{x+2}{5}=\frac{1}{x-2}\left(ĐK:x\ne2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^2-4=5\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{9}\\x=-\sqrt{9}\end{array}\right.\)
\(\frac{x+2}{5}=\frac{1}{x-2}\)
=> (x + 2).(x - 2) = 5
Mà x + 2 > x - 2
=> \(\left[\begin{array}{nghiempt}x+2=5;x-2=1\\x+2=-1;x-2=-5\end{array}\right.\)
+ Với \(\begin{cases}x+2=5\\x-2=1\end{cases}\)\(\Rightarrow x=3\)
+ Với \(\begin{cases}x+2=-1\\x-2=-5\end{cases}\)\(\Rightarrow x=-3\)
Vậy \(\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\)
Ta có: \(\frac{x+2}{5}=\frac{1}{x-2}\)
\(\Rightarrow\left(x+2\right)\left(x-2\right)=5.1\)
\(\Rightarrow x^2-2x+2x-4=5\)
\(\Rightarrow x^2=5-4=1\)
\(\Rightarrow x=1\)
