\(M=-2x^2+2x-3\\ \Leftrightarrow2M=-4x^2+4x-6\\ \Leftrightarrow2M=-\left(4x^2-4x+4\right)-2\\ \Leftrightarrow2M=-\left(2x-2\right)^2-2\\ \Leftrightarrow M=\dfrac{-\left(2x-2\right)^2-2}{2}\)
Ta có :
\(-\left(2x-2\right)^2\le0\\ \Rightarrow-\left(2x-2\right)^2-2\le-2\\ \Rightarrow\dfrac{-\left(2x-2\right)^2-2}{2}\le\dfrac{-2}{2}\\ \Rightarrow M\le-1\)
\(\Rightarrow Max\left(M\right)=-1\Leftrightarrow2x-2=0\Rightarrow x=1\)
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\(N=3x-x^2-4\\ \Leftrightarrow N=-\left(x^2-3x+4\right)\\ \Leftrightarrow N=-\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}+\dfrac{16}{4}\right)\\ \Rightarrow N=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Ta có :
\(-\left(x-\dfrac{3}{2}\right)^2\le0\\ \Rightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\le0+\dfrac{7}{4}\\ \Rightarrow N\le\dfrac{7}{4}\\ \Rightarrow Max\left(M\right)=\dfrac{7}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(P=\dfrac{3}{x^2-6x+10}=\dfrac{3}{\left(x-3\right)^2+1}\)
Ta có :
\(\left(x-3\right)^2\ge0\\ \Rightarrow\left(x-3\right)^2+1\ge1\\ \Rightarrow\dfrac{3}{\left(x-3\right)^2+1}\ge\dfrac{3}{1}\Rightarrow P\ge3\\ \Rightarrow Min\left(P\right)=3\Leftrightarrow x-3=0\Rightarrow x=3\)
