Bài 1:
\(A=\left|x-2\right|+\left|x+y-5\right|+3\)
Ta thấy: \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|x+y-5\right|\ge0\end{matrix}\right.\)\(\forall x,y\)
\(\Rightarrow\left|x-2\right|+\left|x+y-5\right|\ge0\forall x,y\)
\(\Rightarrow\left|x-2\right|+\left|x+y-5\right|+3\ge3\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|x+y-5\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\x+y-5=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Bài 2:
\(B=\dfrac{10}{\left|x+3\right|+\left|y+7\right|+2}\)
Ta thấy: \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|y+7\right|\ge0\end{matrix}\right.\)\(\forall x,y\)
\(\Rightarrow\left|x+3\right|+\left|y+7\right|\ge0\forall x,y\)
\(\Rightarrow\left|x+3\right|+\left|y+7\right|+2\ge2\forall x,y\)
\(\Rightarrow\dfrac{1}{\left|x+3\right|+\left|y+7\right|+2}\le\dfrac{1}{2}\forall x,y\)
\(\Rightarrow B=\dfrac{10}{\left|x+3\right|+\left|y+7\right|+2}\le\dfrac{10}{2}=5\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left|x+3\right|=0\\\left|y+7\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\y+7=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=-7\end{matrix}\right.\)
1/ Vì: \(\left|x-2\right|\ge0\forall x\Rightarrow Min_{\left|x-2\right|}=0\Leftrightarrow x=2\)(1)
Lại có: \(\left|x+y-5\right|\ge0\forall x,y\)
hay \(\left|2+y-5\right|\ge0\forall x,y\)
\(\Rightarrow Min_{\left|2+y-5\right|}=0\Leftrightarrow y=3\) (2)
Từ (1), (2)
\(\Rightarrow MIN_A=3\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
2/ Để \(\dfrac{10}{2+\left|x+3\right|+\left|y+7\right|}\) lớn nhất
\(\Rightarrow2+\left|x+3\right|+\left|y+7\right|\) nhỏ nhất
Ta có: \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\forall x\\\left|y+7\right|\ge0\forall y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}Min_{\left|x+3\right|}=0\Leftrightarrow x=-3\\Min_{\left|y+7\right|}=0\Leftrightarrow y=-7\end{matrix}\right.\)
\(\Rightarrow Min_{2+\left|x+3\right|+\left|y+7\right|}=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-7\end{matrix}\right.\)
\(\Rightarrow MAX_{\dfrac{10}{2+\left|x+3\right|+\left|y+7\right|}}=\dfrac{10}{2}=5\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-7\end{matrix}\right.\)
