a) \(P=\dfrac{\sqrt[]{x}-1}{\sqrt[]{x}+3}\left(x\ge0\right)\)
\(\Leftrightarrow P=\dfrac{\sqrt[]{x}+3-4}{\sqrt[]{x}+3}\)
\(\Leftrightarrow P=1-\dfrac{4}{\sqrt[]{x}+3}\)
Ta có :
\(\sqrt[]{x}\ge0\)
\(\Leftrightarrow\sqrt[]{x}+3\ge3\)
\(\Leftrightarrow\dfrac{1}{\sqrt[]{x}+3}\le\dfrac{1}{3}\)
\(\Leftrightarrow-\dfrac{4}{\sqrt[]{x}+3}\ge-\dfrac{4}{3}\)
\(\Leftrightarrow P=1-\dfrac{4}{\sqrt[]{x}+3}\ge1-\dfrac{4}{3}=-\dfrac{1}{3}\)
Vậy \(GTNN\left(P\right)=-\dfrac{1}{3}\left(khi.x=0\right)\)
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