Chọn A

TH 1: $x\le 1$. Phương trình

$\iff \frac{x^2}{2}-2x+\frac{3}{2}+\frac{x^2}{2}-3x+4=\frac{3}{4}\iff x^2-5x+\frac{19}{4}=0\iff \left[\begin{array}{l}x=\frac{5+\sqrt{6}}{2}\left(L\right)\\ x=\frac{5-\sqrt{6}}{2}\left(L\right)\end{array}\right.$.

TH 2: $1<x<2$. Phương trình $\iff -\frac{x^2}{2}+2x-\frac{3}{2}+\frac{x^2}{2}-3x+4=\frac{3}{4}\iff x=\frac{7}{4}(t/m)$.

TH 3: $2\le x\le 3$. Ta được $-\frac{x^2}{2}+2x-\frac{3}{2}-\frac{x^2}{2}+3x-4=\frac{3}{4}\iff -x^2+5x-\frac{25}{4}=0\iff x=\frac{5}{2}(t/m)$

TH 4: $3<x<4$. Phương trình $\iff \frac{x^2}{2}-2x+\frac{3}{2}-\frac{x^2}{2}+3x-4=\frac{3}{4}\iff x=\frac{13}{4}(t/m)$.

TH 5: $x\ge 4$. Phưong trình

$\iff \frac{x^2}{2}-2x+\frac{3}{2}+\frac{x^2}{2}-3x+4=\frac{3}{4}\iff x^2-5x+\frac{19}{4}=0\iff \left[\begin{array}{l}x=\frac{5+\sqrt{6}}{2}\left(L\right)\\ x=\frac{5-\sqrt{6}}{2}\left(L\right)\end{array}\right.$.

Vậy nghiệm của phương trình là $x=\frac{7}{4},x=\frac{5}{2},x=\frac{13}{4}$.