Question

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)

Answer 1

ĐK: `x \ne 0;2`

`(x+2)/(x-2)-2/(x^2-2x)=1/x`

`<=>x(x+2)-2=x-2`

`<=> x^2+2x-2=x-2`

`<=>x^2+x=0`

`<=>` \(\left[{}\begin{matrix}x=0\left(L\right)\\x=-1\end{matrix}\right.\)

Vậy `S={-1}`.

Answer 2

Đk x khác 0, x khác 2 

\(\dfrac{x\left(x+2\right)-2-\left(x-2\right)}{x\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+2x-2-x+2}{x\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+x}{x\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{x\left(x-2\right)}=0\Leftrightarrow\dfrac{x+1}{x-2}=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)