\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne-1\end{cases}}\)
\(\frac{x-2}{x+3}=\frac{x-3}{x+1}\)
\(\Rightarrow\left(x-2\right)\left(x+1\right)=\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-x-2=x^2-9\)
\(\Leftrightarrow-x-2=-9\)
\(\Leftrightarrow x=7\)( thỏa mãn ĐKXĐ )
Vậy \(x=7\)
\(\frac{x-2}{x+3}=\frac{x-3}{x+1}\)
\(\left(x-2\right)\left(x+1\right)=\left(x+3\right)\left(x-3\right)\)
\(x^2-x-2=x^2-9\)
\(x^2-x=x^2-9+2\)
\(x^2-x=x^2-7\)
\(-x=x^2-7-x^2\)
\(-x=-7\)
\(x=7\)
\(\frac{x-2}{x+3}=\frac{x-3}{x+1}\)ĐK : x khác -3 ; -1
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=\left(x+3\right)\left(x-3\right)\)
\(\Leftrightarrow x^2+x-2x-2=x^2-9\Leftrightarrow x^2-x-2=x^2-9\)
\(\Leftrightarrow-x=-7\Leftrightarrow x=7\)( tm )
