\(3x-2x^2+1=\frac{3}{2}x+\frac{3}{2}x-2x^2-\frac{9}{8}+\frac{17}{8}=\left(-2x^2+\frac{3}{2}x\right)+\left(\frac{3}{2}x-\frac{9}{8}\right)+\frac{17}{8}\)
\(=-2\left(x-\frac{3}{4}\right)+\frac{3}{2}\left(x-\frac{3}{4}\right)+\frac{17}{8}\)
\(=\left(x-\frac{3}{4}\right)\left(-2x+\frac{3}{2}\right)+\frac{17}{8}=\left(x-\frac{3}{4}\right).\left(-2\right)\left(x-\frac{3}{4}\right)+\frac{17}{8}\)
\(=-2.\left(x-\frac{3}{4}\right)^2+\frac{17}{8}\)
Do \(\left(x-\frac{3}{4}\right)^2>=0và-2
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