K=\(\frac{x^2-3x+2x-6}{x^2-4}\)
=\(\frac{x\left(x-3\right)+2\left(x-3\right)}{x^2-4}\)
=\(\frac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)
=\(\frac{x-3}{x-2}\)
ta có K=3
\(\Rightarrow3=\frac{x-3}{x-2}\)
\(\Leftrightarrow3x-6=x-3\)
\(\Leftrightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)=1.5
vây x=1,5