Bài giải:
1.
$(2x-1)(3-2x)+4=x^2-3x-1$
<=>$6x-4x^2-3+2x+4=x^2-3x-1$
<=>$-5x^2+11x+2=0$
<=>\(\left[ \begin{array}{l}x_{1}=\frac{11+\sqrt{161}}{10}\\x_{2}=\frac{11-\sqrt{161}}{10}\end{array} \right.\)
2.
$(3x-5)(x-2)=-x(x-2)-5$
<=>$3x^2-6x-5x+10=-x^2+2x-5$
<=>$4x^2-13x+15=0$
Phương trình vô nghiệm
3.
$x(x+3)=15-(3x-1)$
<=>$x^2+3x=15-3x+1$
<=>$x^2+6x-16=0$
<=>\(\left[ \begin{array}{l}x_{1}=2\\x_{2}=-8\end{array} \right.\)
