a.
Ta có: \(\left(x-1\right)^2\ge0\forall x\\ \Rightarrow\left(x-1\right)^2-\frac{1}{3}\ge-\frac{1}{3}\forall x\)
Vậy \(A_{Min}=-\frac{1}{3}\) \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
a) \(A=-\frac{1}{3}+\left(x-1\right)^2\)
Ta có: \(\left(x-1\right)^2\ge0\) với mọi \(x\)
\(\Rightarrow-\frac{1}{3}+\left(x-1\right)^2\ge-\frac{1}{3}\) với mọi \(x\)
\(\Leftrightarrow A\ge-\frac{1}{3}\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\) \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy \(MinA=-\frac{1}{3}\Leftrightarrow x=1\).
b) \(B=5-2\left(3x-1\right)^4\)
Ta có: \(\left(3x-1\right)^4\ge0\) với mọi \(x\)
\(\Leftrightarrow-2\left(3x-1\right)^4\le0\) với mọi \(x\)
\(\Rightarrow5-2\left(3x-1\right)^4\le5\) với mọi \(x\)
\(\Leftrightarrow B\le5\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left(3x-1\right)^4=0\Leftrightarrow3x-1=0\Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\)
Vậy \(MaxB=5\Leftrightarrow x=\frac{1}{3}\).
c) \(C=\left(x+1\right)^2+\left|y-5\right|-2\)
Ta có: \(\left(x+1\right)^2\ge0\) với mọi \(x\)
\(\left|y-5\right|\ge0\) với mọi \(y\)
\(\Rightarrow\left(x+1\right)^2+\left|y-5\right|-2\ge-2\) với mọi \(x,y\)
\(\Leftrightarrow C\ge-2\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left|y-5\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=5\end{matrix}\right.\)
Vậy \(MinC=-2\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=5\end{matrix}\right.\).
