Question

Tìm GTLN

M = 2x -2 -3x²

P = 2- x² - y² -2(x+y)

Giup mình với nha !

Answer 1

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 \(P_{\max}=4\Leftrightarrow (x+1)^2=(y+1)^2=0\Leftrightarrow x=y=-1\)