Giải thích các bước giải:
$\eqalign{ & |2{x^2} - 5x + 4| = 2x - 1 \cr & Ta\,co:\,2{x^2} - 5x + 4 \cr & = 2({x^2} - \frac{5}{2}x + \frac{{25}}{{16}}) + \frac{7}{8} \cr & = 2{(x - \frac{5}{4})^2} + \frac{7}{8} \geqslant \frac{7}{8}\forall x \cr & \Rightarrow |2{x^2} - 5x + 4| = 2{x^2} - 5x + 4 \cr & Khi\,đó:\,2{x^2} - 5x + 4 = 2x - 1 \cr & \Leftrightarrow 2{x^2} - 7x + 5 = 0 \cr & \Leftrightarrow (2x - 5)(x - 1) = 0 \cr & \Leftrightarrow 2x - 5 = 0\,hoặc\,x - 1 = 0 \cr & \Leftrightarrow x = \frac{5}{2}\,hoặc\,x = 1 \cr} $
$\eqalign{ & 3{x^2} + x - 4|x + 2| + 8 = 0 \cr & + )\,Nếu\,x > - 2 \cr & \Rightarrow x + 2 > 0 \cr & \Rightarrow |x + 2| = x + 2 \cr & Khi\,đó:\,3{x^2} + x - 4(x + 2) + 8 = 0 \cr & \Leftrightarrow 3{x^2} - 3x = 0 \cr & \Leftrightarrow 3x(x - 1) = 0 \cr & \Leftrightarrow x = 0\,hoac\,x - 1 = 0 \cr & \Leftrightarrow x = 0\,hoac\,x = 1(tm\,ĐKXĐ) \cr & + )\,Nếu\,x \leqslant - 2 \cr & \Rightarrow x + 2 \leqslant 0 \cr & \Rightarrow |x + 2| = - x - 2 \cr & Khi\,đó:\,3{x^2} + x - 4( - x - 2) + 8 = 0 \cr & \Leftrightarrow 3{x^2} + 5x + 16 = 0 \cr & \Leftrightarrow 3({x^2} + \frac{5}{3}x + \frac{{25}}{{36}}) + \frac{{167}}{{12}} = 0 \cr & \Leftrightarrow 3{(x + \frac{5}{6})^2} + \frac{{167}}{{12}} = 0 \cr & Vi\,{(x + \frac{5}{6})^2} \geqslant 0\forall x \cr & \Rightarrow 3{(x + \frac{5}{6})^2} + \frac{{167}}{{12}} > 0 \cr & \Rightarrow pt\,vô\,nghiệm \cr} $
Vậy pt có nghiệm x=0 hoặc x=1
