Đáp án:
$MaxA=1$
$MinA=\dfrac{-1}{2}$
Giải thích các bước giải:
+$A=\dfrac{4x+1}{4x^2+2}$
$\rightarrow 1-A=1-\dfrac{4x+1}{4x^2+2}=\dfrac{4x^2-4x+1}{4x^2+2}=\dfrac{(2x-1)^2}{4x^2+2}\ge 0$
$\rightarrow A\le 1\rightarrow MaxA=1$
+$A+\dfrac{1}{2}=\dfrac{4x+1}{4x^2+2}+\dfrac{1}{2}=\dfrac{4x+1+2x^2+1}{4x^2+2}=\dfrac{2(x+1)^2}{4x^2+2}\ge 0$
$\rightarrow A\ge \dfrac{-1}{2}$
$\rightarrow MinA=\dfrac{-1}{2}$
