ta thấy\(\sqrt{\left(2x+1\right)^2+4}\ge0\forall x\)
3/4y\(^2\)-1\(\ge0\forall x\)
suy ra \(\sqrt{\left(2x+1\right)^2+4}+3\)/4y\(^2\)-1/\(\ge0\forall x,y\)
=>min a=5
dau =xảy ra <=>x=\(\dfrac{3}{2}\),y=\(\dfrac{1}{2}\)
\(A=\sqrt{\left(2x+1\right)^2+4}+3\left|4y^2-1\right|+5\)
Ta thấy: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2+4\ge4\forall x\)
\(\Rightarrow\sqrt{\left(2x+1\right)^2+4}\ge\sqrt{4}=2\forall x\)
Lại có: \(\left|4y^2-1\right|\ge0\forall y\Rightarrow3\left|4y^2-1\right|\ge0\forall y\)
\(\Rightarrow\sqrt{\left(2x+1\right)^2+4}+3\left|4y^2-1\right|\ge2\forall x,y\)
\(\Rightarrow A=\sqrt{\left(2x+1\right)^2+4}+3\left|4y^2-1\right|+5\ge7\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(A_{Min}=7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
