a, \(A=\left(1+x^2\right)+\left|y+\frac{1}{5}\right|-3\)
\(A=\left|y+\frac{1}{5}\right|+x^2-2\)
Ta có: \(\begin{cases}\left|y+\frac{1}{5}\right|\ge0\\x^2\ge0\end{cases}\)\(\Rightarrow\left|y+\frac{1}{5}\right|+x^2\ge0\Rightarrow\left|y+\frac{1}{5}\right|+x-2\ge-2\Rightarrow A\ge-2\)
Vậy GTNN của \(A=\left(1+x^2\right)+\left|y+\frac{1}{5}\right|-3\) là - 2.
b, a đâu bạn ?