P có nghĩa khi x\(\ne\)4
\(P=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\right]\left(x-4\right)\)
\(P=\left(\dfrac{x-2\sqrt{x}+\sqrt{x}+2+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)\)
\(P=x-2\sqrt{x}+\sqrt{x}+2+1=x-3\sqrt{x}+3\)
\(a,ĐK:x\ge0;x\ne4\\ P=\dfrac{x-2\sqrt{x}+\sqrt{x}+2+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\\ P=x-\sqrt{x}+3\\ b,x=\dfrac{1}{\sqrt{5}-2}-\sqrt{5}+14=\sqrt{5}+2-\sqrt{5}+14\\ x=16\\ \Leftrightarrow P=16-4+3=15\\ c,P=x-\sqrt{x}+3=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{11}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\\ P_{min}=\dfrac{11}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
