Tìm GTNN : $B = \dfrac{1}{3}|x-2|+2|3 – \dfrac{1}{2}y|+4$ Tìm GTLN : $B = \dfrac{1}{3+\dfrac{1}{2}|2x -3|}$

Đáp án:

a) $Min_{B}=4\Leftrightarrow \left\{\begin{matrix}
x=2\\ 
y=6
\end{matrix}\right.$

b) $Max_{B}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}$

Giải thích các bước giải:

a) Ta có: $\dfrac{1}{3}\left | x-2 \right |+2\left | 3-\dfrac{1}{2}y \right |\geq 0\forall x,y$
$\rightarrow \dfrac{1}{3}\left | x-2 \right |+2\left | 3-\dfrac{1}{2}y \right |+4\geq 4\forall x,y$
$\rightarrow B\geq 4\forall x,y$
Dấu “=” xảy ra khi $\left\{\begin{matrix}
x-2=0\\ 
3-\dfrac{1}{2}y=0
\end{matrix}\right.$
$\rightarrow \left\{\begin{matrix}
x=2\\ 
y=6
\end{matrix}\right.$
Vậy $Min_{B}=4\Leftrightarrow \left\{\begin{matrix}
x=2\\ 
y=6
\end{matrix}\right.$
b) Ta có: $\dfrac{1}{2}\left | 2x-3 \right |\geq 0\forall x$
$\rightarrow 3+\dfrac{1}{2}\left | 2x-3 \right |\geq 3\forall x$
$\rightarrow \dfrac{1}{3+\dfrac{1}{2}\left | 2x-3 \right |}\leq \dfrac{1}{3}\forall x$
$\rightarrow B\leq \dfrac{1}{3}\forall x$
Dấu “=” xảy ra khi $2x-3=0\rightarrow x=\dfrac{3}{2}$
Vậy $Max_{B}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}$