a ) \(2x^2-5x+4\)
\(=2\left(x^2-\dfrac{5}{2}x+2\right)\)
\(=2\left(x^2-2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{7}{16}\right)\)
\(=2\left[\left(x-\dfrac{5}{4}\right)^2+\dfrac{7}{16}\right]\)
\(=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{7}{8}\)
Do\(2\left(x-\dfrac{5}{4}\right)^2\ge0\forall x\Rightarrow2\left(x-\dfrac{5}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}>0\left(đpcm\right)\)
b ) \(-x^2+4x-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-4x+4+1\right)\)
\(=-\left[\left(x-2\right)^2+1\right]\)
\(=-\left(x-2\right)^2-1\)
Do \(-\left(x-2\right)^2\le0\forall x\Rightarrow-\left(x-2\right)^2-1\le-1< 0\left(đpcm\right)\)
c ) Sai đề : Đây là đề theo cách sửa của mik :
\(-4+3x-3x^2\)
\(=-3\left(x^2-x+\dfrac{4}{3}\right)\)
\(=-3\left(x^2-x+\dfrac{1}{4}+\dfrac{13}{12}\right)\)
\(=-3\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{13}{12}\right]\)
\(=-3\left(x-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\)
Do \(-3\left(x-\dfrac{1}{2}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\le\dfrac{-13}{4}< 0\left(đpcm\right)\)
