Question

[9.Nguyễn Phúc Khang 8/2]

Cứu mình với ạ

 

Answer 1

[Nguyễn Ngọc Huy Toàn]

\(ĐK:x\ne0;2\)

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

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

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

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

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

\(\Leftrightarrow x\left(x+1\right)=0\)

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

Answer 2

[nhung mai]

\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)       ĐKXĐ: x≠0, x≠2

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

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

<=>x\(^2\) + 2x - x = 2-2

<=> x\(^2\) + x =0

<=> x(x + 1) = 0

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

mà ĐKXĐ: x≠0, x≠2

Vậy x = -1