\(a,A=\left|x-5\right|+x^2+8xy+16y^2+17\\ A=\left|x-5\right|+\left(x+4y\right)^2+17\ge17\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}x-5=0\\x+4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(MinA=17\Leftrightarrow\left(x;y\right)=\left(5;-\dfrac{5}{4}\right)\)
\(b,B=\left(x+2y\right)^{2020}+y^2+14y+55\\ B=\left(x+2y\right)^{2020}+\left(y+7\right)^2+6\ge6\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}x+2y=0\\y+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=14\\y=-7\end{matrix}\right.\)
Vậy \(MinB=6\Leftrightarrow\left(x;y\right)=\left(14;-7\right)\)
a.\(A=\left|x-5\right|+x^2+8xy+16y^2+17\)
\(A=\left|x-5\right|+\left(x+4y\right)^2+17\)
Ta có: \(\left\{{}\begin{matrix}x-5\ge0\\x+4y\ge0\end{matrix}\right.\)
\(\rightarrow Min_A=17\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=5\\5+4y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(Min_A=17\) khi \(\left\{{}\begin{matrix}x=5\\y=-\dfrac{5}{4}\end{matrix}\right.\)
b.\(B=\left(x+2y\right)^{2020}+y^2+14y+55\)
\(B=\left(x+2y\right)^{2020}+\left(y+7\right)^2+6\)
Ta có: \(\left\{{}\begin{matrix}\left(x+2y\right)^{2020}\ge0\\\left(y+7\right)^2\ge0\end{matrix}\right.\)
`->` \(Min_B=6\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=14\\y=-7\end{matrix}\right.\)
Vậy \(Min_B=6\) khi \(\left\{{}\begin{matrix}x=14\\y=-7\end{matrix}\right.\)