a, Với \(x\ge0,x\ne1\)
\(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Với \(x\ge0,x\ne1\)
Để \(P=\dfrac{2}{7}\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{7}\)
\(\Rightarrow x+\sqrt{x}+1=7\)
\(\Leftrightarrow x+\sqrt{x}-6=0\)
\(\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-2\right)+3.\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right).\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(l,do\sqrt{x}\ge0\right)\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow x=4\) (t/m)
Vậy x = 4 thì 'P = 2/7'
