\(P=3-4x-x^2=-\left(x^2+4x+4\right)+7\)
\(P=-\left(x+2\right)^2+7\)
\(Do-\left(x+2\right)^2\le0\Leftrightarrow P\le7\)
Dấu "=" xảy ra khi x + 2 =0
=> x = -2
Vậy Max P = 7 khi x = - 2
b) \(Q=2x-2-3x^2\)
\(\Leftrightarrow Q=-\left[\left(3x^2-2x+\frac{1}{3}\right)+\frac{5}{3}\right]\)
\(\Leftrightarrow Q=-\left[\left(\sqrt{3}x-\frac{1}{\sqrt{3}}\right)^2+\frac{5}{3}\right]\le\frac{5}{3}\)
Dấu " = " xảy ra :
\(\Leftrightarrow\sqrt{3}x-\frac{1}{\sqrt{3}}=0\)
\(\Leftrightarrow\sqrt{3}x=\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy \(Max_Q=\frac{5}{3}\Leftrightarrow x=\frac{1}{3}\)
c) \(R=2-x^2-y^2-2\left(x+y\right)\)
\(\Leftrightarrow R=-\left[\left(x^2+2x+1\right)+\left(y^2+2y+1\right)-4\right]\)
\(\Leftrightarrow R=-\left[\left(x+1\right)^2+\left(y+1\right)^2-4\right]\le-4\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x+1=0\\y+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
Vậy \(Max_Q=-4\Leftrightarrow x=y=-1\)
d) \(S=-x^2+4x-9\)
\(\Leftrightarrow S=-\left[\left(x^2-4x+4\right)+5\right]\)
\(\Leftrightarrow S=-\left[\left(x-2\right)^2+5\right]\le5\)
Dấu " = " xảy ra :
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_S=5\Leftrightarrow x=2\)
