a) \(\dfrac{2x-1}{x-3}-\dfrac{3}{x-2}=2\) ĐK: x\(\ne3;x\ne2\)
<=>\(\dfrac{\left(2x-1\right).\left(x-2\right)}{\left(x-3\right).\left(x-2\right)}-\dfrac{3.\left(x-3\right)}{\left(x-3\right).\left(x-2\right)}-\dfrac{2.\left(x-2\right).\left(x-3\right)}{\left(x-2\right).\left(x-3\right)}=0\)
=>\(\left(2x-1\right).\left(x-2\right)-3x+9-\left(2x-4\right).\left(x-3\right)=0\)
<=>\(2x^2-x-4x+2-3x+9-\left(2x^2-4x-6x+12\right)=0\)
<=>\(2x^2-8x+11-2x^2+4x+6x-12=0\)
<=>\(2x=1\)
<=>\(x=\dfrac{1}{2}\) (TMĐK)
Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{1}{2}\right\}\)
b) \(x^2-4+x.\left(x-2\right)=0\)
<=>\(\left(x-2\right).\left(x+2\right)+x.\left(x-2\right)=0\)
<=>\(\left(x-2\right).\left(x+2+x\right)=0\)
<=>\(\left(x-2\right).\left(2x+2\right)=0\)
<=>\(\left(x-2\right).2.\left(x+1\right)=0\)
<=>\(\left(x-2\right).\left(x+1\right)=0\) (vì 2>0)
<=>\(\left[{}\begin{matrix}\left(x-2\right)=0\\\left(x+1\right)=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S=\(\left\{2;-1\right\}\)