a,\(A=\left(x-2\right)^2+3\ge3\forall x\\ \)
Dấu bằng xảy ra khi
\(x-2=0\\ x=2\)
Vậy \(Min_A=3khix=2\)
\(b,B=\left(x-4\right)^2-16\ge-16\forall x\)
Dấu bằng xảy ra khi \(x-4=0=>x=4\)
Vậy \(Min_B=-16khix=4\)
\(c,C=-2\left(x-2\right)^2-7\le-7\forall x\)
Dấu bằng xảy ra khi
\(x-2=0=>x=2\)
Vậy \(Max_C=-7khix=2\)
\(d,D=-\left(x-3\right)^2-\left(y-1\right)^2+2\le2\forall x,y\)
Dấu bằng xảy ra khi
\(\left\{{}\begin{matrix}x-3=0\\y-1=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy \(max_D=2khix=3;y=1\)
\(e,E=\left(x^2+1\right)^2+4\ge4\forall x\)
Dấu bằng xảy ra khi
\(x^2+1=0\left(voli\right)\left(x^2+1\ge0\forall x\right)\)
Vậy \(Min_E=4\forall x\)
\(A=\left(x-2\right)^2+3\ge3\)
Dấu bằng xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(MinA=3\Leftrightarrow x=2\)
\(B=\left(x-4\right)^2-16\ge-16\)
Dấu bằng xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy \(MinB=-16\Leftrightarrow x=4\)
\(C=-2\left(x-2\right)^2-7\le-7\)
Dấu bằng xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(MaxC=-7\Leftrightarrow x=2\)
\(D=-\left(x-3\right)^2-\left(y-1\right)^2+2\le2\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy \(MaxD=2\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
\(E=\left(x^2+1\right)^2+4\ge\left(0+1\right)^2+4=1+4=5\)
Dấu bằng xảy ra \(\Leftrightarrow x=0\)
Vậy \(MinE=5\Leftrightarrow x=0\)
