1)(2x-1)(3-2x)+4=x bình-3x-1 2)(3x-5)(x-2)=-x(x-2)-5 3)x(x+3)=15-(3x-1)

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.

$x(x+3)=15-(3x-1)$

<=>$x^2+3x=15-3x+1$

<=>$x^2+6x-16=0$

<=>\(\left[ \begin{array}{l}x_{1}=2\\x_{2}=-8\end{array} \right.\)