-ĐKXĐ: \(x\ne\pm2\).
\(\dfrac{x}{x-2}-\dfrac{x-1}{x+2}=\dfrac{3}{x^2-4}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x-\left(x^2-3x+2\right)=3\)
\(\Leftrightarrow x^2+2x-x^2+3x-2=3\)
\(\Leftrightarrow5x=5\)
\(\Leftrightarrow x=1\left(nhận\right)\)
-Vậy \(S=\left\{1\right\}\)
Đúng 1
Bình luận (0)
