Question

giải giúp mik bài này ạ \(\dfrac{x+2}{x-2}\) -\(\dfrac{1}{x}\) = \(\dfrac{2}{x^2-2x}\)

Answer 1

<=> \(\dfrac{x+2}{x-2}\)-\(\dfrac{1}{x}\)=\(\dfrac{2}{x\left(x-2\right)}\)

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

ok, ở đây đã có mẫu chung rồi, em cứ vậy làm tiếp thôi :D

Answer 2

\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (ĐKXĐ: \(x\ne0;x\ne2\))

\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x+2-2=0\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow S=\left\{-1\right\}\)

Answer 3

\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)(ĐKXĐ: \(x\ne2,x\ne0\))

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

\(\Leftrightarrow x=-1\) (thỏa mãn điều kiện)

Vậy \(x=-1\) là nghiệm của phương trình