Vì \(\left(2x+\frac{1}{4}\right)^4\ge0;\left|y+\frac{11}{3}\right|\ge0\)
Suy ra:\(\left(2x+\frac{1}{4}\right)^4\ge0;\left|y+\frac{11}{3}\right|-1\ge-1\)
Vậy dấu = xảy ra khi \(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{4}=0\\y+\frac{11}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{8}\\y=-\frac{11}{3}\end{cases}}\)
Min A=-1 khi \(x=-\frac{1}{8};y=-\frac{11}{3}\)
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