Question

\(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

Answer 1

Ta có \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

<=> \(\dfrac{3\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow3\left(x-1\right)-\left(x+3\right)=\left(x+1\right)\left(x-3\right)\)

\(\Leftrightarrow3x-3-x-3=x^2-2x-3\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\)(x-3)(x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy....