Question

Em đang gấp , giúp vs ạ 

Answer 1

Ta có: \(\left|x+\dfrac{1}{7}\right|\ge0\forall x,\left|y-12\right|\ge0\forall y\)

\(\Rightarrow A=\left|x+\dfrac{1}{7}\right|+\left|y-12\right|\ge0\)

\(minA=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x+\dfrac{1}{7}=0\\y-12=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{7}\\y=12\end{matrix}\right.\)

Answer 2

\(A=\left|x+\dfrac{1}{7}\right|+\left|y-12\right|\)

Vì \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{7}\right|\text{≥}0\\\left|y-12\right|\text{≥}0\end{matrix}\right.\)

⇒\(\left|x+\dfrac{1}{7}\right|+\left|y-12\right|\text{≥}0\)

Min \(A=0\)⇔\(\left\{{}\begin{matrix}x+\dfrac{1}{7}=0\\y-12=0\end{matrix}\right.\)

                 ⇔\(\left\{{}\begin{matrix}x=-\dfrac{1}{7}\\y=12\end{matrix}\right.\)