Câu hỏi của Dang Le Tu Quynh

Question

$\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}$

giai giup minh

svtkvtm · Accepted Answer

$ĐK:x\ne2;x\ne0$

$\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\Leftrightarrow\frac{\left(x+2\right)x}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\Leftrightarrow\frac{x^2+2x-x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\Leftrightarrow x^2+x+2=2\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\x+1=0\Leftrightarrow x=-1\left(thoảmanx\right)\end{matrix}\right..Vậy:x=-1$Câu trả lời: $ĐK:x\ne2;x\ne0$

$\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\Leftrightarrow\frac{\left(x+2\right)x}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\Leftrightarrow\frac{x^2+2x-x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\Leftrightarrow x^2+x+2=2\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\x+1=0\Leftrightarrow x=-1\left(thoảmanx\right)\end{matrix}\right..Vậy:x=-1$

B.Thị Anh Thơ · Answer

$\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}$

$\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)},x\ne2,x\ne0$

$\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\left(x-2\right)}=0$

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

$\frac{x^2+x}{x\left(x-2\right)}=0$

$\frac{x\left(x-1\right)}{x\left(x-2\right)}=0$

$\frac{x+1}{x-2}=0$

x+1=0

x=-1(TM)Câu trả lời: $\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}$