Question

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Answer 1

a) \(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)

\(\Rightarrow P=\left(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)

\(\Rightarrow P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(\Rightarrow P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(\Rightarrow P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(\Rightarrow P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(\Rightarrow P=\dfrac{2}{x+\sqrt{x}+1}\)

Answer 2

b) \(P=\dfrac{2}{7}\\ \Rightarrow\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{7}\\ \Rightarrow x+\sqrt{x}+1=7\\ \Rightarrow x+\sqrt{x}-6=0\\ \Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-3\left(vôlí\right)\end{matrix}\right.\\ \Rightarrow x=4\)