Question

giải các phương trình sau :

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

Answer 1

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : \(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{2}{x\left(x+1\right)}\)

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

`=> x(x-1) -(x+1)=2`

`<=>x^2 - x -x-1=2`

`<=> x^2 -2x-1+2=0`

`<=> x^2 -2x +1=0`

`<=> (x+1)^2=0`

`<=>x+1=0`

`<=>x=-1(ktm)`

Vậy pt vô nghiệm 

Answer 2