Lời giải:
\(M=2x-2-3x^2=\frac{-5}{3}-(3x^2-2x+\frac{1}{3})\)
\(=-\frac{5}{3}-3(x^2-\frac{2}{3}x+\frac{1}{3^2})\)
\(=-\frac{5}{3}-3(x-\frac{1}{3})^2\)
Vì \((x-\frac{1}{3})^2\geq 0, \forall x\in\mathbb{R}\)
\(\Rightarrow M=-\frac{5}{3}-3(x-\frac{1}{3})^2\leq -\frac{5}{3}-3.0=-\frac{5}{3}\)
Vậy \(M_{\max}=\frac{-5}{3}\Leftrightarrow x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\)
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\(P=2-x^2-y^2-2(x+y)=4-[x^2+y^2+2(x+y)+2]\)
\(=4-[(x^2+2x+1)+(y^2+2y+1)]\)
\(=4-[(x+1)^2+(y+1)^2]\)
Vì \((x+1)^2\geq 0; (y+1)^2\geq 0, \forall x,y\)
\(\Rightarrow (x+1)^2+(y+1)^2\geq 0\)
\(\Rightarrow P=4-[(x+1)^2+(y+1)^2]\leq 4-0=4\)
Vậy \(P_{\max}=4\Leftrightarrow (x+1)^2=(y+1)^2=0\Leftrightarrow x=y=-1\)
