a, \(P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{x-4}{2}=\dfrac{2\sqrt{x}\left(x-4\right)}{2\left(x-4\right)}=\sqrt{x}\) b, Với \(x\ge0,x\ne4\) , để \(P=6-\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=6-\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\)
Vậy với x=9 thì \(P=6-\sqrt{x}\)
c, \(Q=x-P=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy MinQ=-1/4, đạt đc khi x=1/4
