\(A=\left|x+3\right|+\left(y-1\right)^{2018}-4\)
Vì \(\left|x+3\right|\)và \(\left(y-1\right)^{2018}\)\(\ge0\forall x;y\)
\(\Rightarrow A\ge4\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy.....
\(C=4-\left|3x-5\right|-\left|5y+8\right|\)
\(C=4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\)
Lí luận như câu a) ta có :
\(C\le4\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-8}{5}\end{cases}}\)
Vậy,...........
\(A=\left|x+3\right|+\left(y-1\right)^{2018}-4\)
Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-1\right)^{2018}\ge0\forall y\end{cases}}\)
\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}-4\ge-4\forall x;y\)
\(A=-4\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-1\right)^{2018}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)
Vậy \(A_{min}=-4\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
\(C=4-\left|3x-5\right|-\left|5y+8\right|\)
Ta có: \(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left|5y+8\right|\ge0\forall y\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}-\left|3x-5\right|\le0\forall x\\-\left|5y+8\right|\le0\forall y\end{cases}}\)
\(\Rightarrow4-\left|3x-5\right|-\left|5y+8\right|\le4\forall x;y\)
\(C=4\Leftrightarrow\hept{\begin{cases}-\left|3x-5\right|=0\\-\left|5y+8\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{8}{5}\end{cases}}}\)
Vậy \(C_{max}=4\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{8}{5}\end{cases}}\)
Tham khảo nhé~
Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-1\right)^{2018}\ge0\forall y\end{cases}}\)
\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}\ge0\forall x,y\)
\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}-4\ge-4\forall x,y\)
\(\Rightarrow A\ge-4\)
\(A=-4\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-1\right)^{2018}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)
Vậy MinA=-4\(\Leftrightarrow\)x=-3: y=1
Ta có: \(C=4-\left|3x-5\right|-\left|5y+8\right|\)
\(=4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\)
Vì\(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left|5y+8\right|\ge0\forall y\end{cases}}\)
\(\Rightarrow\left|3x-5\right|+\left|5y+8\right|\ge0\forall x,y\)
\(\Rightarrow-\left(\left|3x-5\right|+\left|5y+8\right|\right)\le0\forall x,y\)
\(\Rightarrow4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\ge4\forall x,y\)
\(\Rightarrow C\ge4\)
\(C=4\Leftrightarrow\hept{\begin{cases}\left|3x-5\right|=0\\\left|5y+8\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-8}{5}\end{cases}}}\)
Vậy MaxC=4\(\Leftrightarrow\)x=\(\frac{5}{3}\): y=\(\frac{-8}{5}\)
