Question

Answer 1

\(\left(\left|x-2\right|+3\right)^2+1\ge10\forall x\)

\(\Leftrightarrow A\le\dfrac{1}{10}\forall x\)

Dấu '=' xảy ra khi x=2

Answer 2

\(A=\dfrac{1}{\left(\left|x-2\right|+3\right)^2+1}\le\dfrac{1}{\left(0+3\right)^2+1}=\dfrac{1}{10}\\ A_{max}=10\Leftrightarrow x-2=0\Leftrightarrow x=2\)