a)\(P=\dfrac{1}{x-4\sqrt{x}+4+1}=\dfrac{1}{\left(\sqrt{x}-2\right)^2+1}\)
ta có: \(\left(\sqrt{x}-2\right)^2\ge0\Rightarrow\left(\sqrt{x}-2\right)^2+1\ge1\)
\(\Rightarrow\dfrac{1}{\left(\sqrt{x}-2\right)^2+1}\le1\)
vậy:MaxP=1khi x=4
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