\(A=2\left(x+3\right)^2-5\)
\(\left(x+3\right)^2\ge0\Rightarrow2\left(x+3\right)^2\ge0\)
\(A_{MIN}\Rightarrow2\left(x+3\right)^2_{MIN}\)
\(2\left(x+3\right)^2_{MIN}=0\)
\(A_{MIN}=0-5=-5\)
\(B=x^4+3x^2+2\)
\(x^4\ge0;x^2\ge0\Rightarrow3x^2\ge0\)
\(B_{MIN}\Rightarrow x^4_{MIN};3x^2_{MIN}\)
\(x^4_{MIN}=0;3x^2_{MIN}=0\)
\(B_{MIM}=0+0+2=2\)
\(C=\left(x^4+5\right)^2\)
\(\left(x^4+5\right)^2\ge0\)
\(C_{MIN}\Rightarrow\left(x^4+5\right)^2_{MIN}\)
\(\left(x^4+5\right)^2_{MIN}=0\)
\(\Rightarrow C_{MIN}=0\)
\(D=\left(x-1\right)^2+\left(y+2\right)^2\)
\(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\)
\(D_{MIN}\Rightarrow\left(x-1\right)^2_{MIN};\left(y+2\right)^2_{MIN}\)
\(\left(x-1\right)^2_{MIN}=0;\left(y+2\right)^2_{MIN}=0\)
\(D_{MIN}=0+0=0\)
a/ Ta có: \(2\left(x+3\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x+3\right)^2-5\ge-5\)
Dấu ''='' xảy ra \(\Leftrightarrow\left(x+3\right)^2=0\Rightarrow x=-3\)
Vậy \(A_{MIN}=-5\Leftrightarrow x=-3\)
b/ Có: \(\left\{{}\begin{matrix}x^4\ge0\\3x^2\ge0\end{matrix}\right.\)\(\forall x\)
\(\Rightarrow x^4+3x^2\ge0\Rightarrow x^4+3x^2+2\ge2\)
Dấu ''='' xảy ra \(\Leftrightarrow x=0\)
Vậy \(B_{MIN}=2\Leftrightarrow x=0\)
c/ Ta có: \(x^4\ge0\forall x\Rightarrow x^4+5\ge5\)
\(\Rightarrow\left(x^4+5\right)^2\ge5^2=25\)
Dấu ''='' xảy ra \(\Leftrightarrow x=0\)
Vậy \(C_{MIN}=25\Leftrightarrow x=0\)
d/ Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\Rightarrow x=1\\\left(y+2\right)^2=0\Rightarrow y=-2\end{matrix}\right.\)
Vậy \(D_{MIN}=0\) khi \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
