Cho \(P = \left( {\frac{1}{{\sqrt x }} + \frac{{\sqrt x }}{{\sqrt x + 1}}} \right):\frac{{\sqrt x }}{{x + \sqrt x }}\).

a) Rút gọn P.

b) Tìm giá trị của P khi x = 4.

c) Tìm x để \(P = \frac{{13}}{3}\).

a) Điều kiện: x > 0

\(P = \left( {\frac{1}{{\sqrt x }} + \frac{{\sqrt x }}{{\sqrt x + 1}}} \right):\frac{{\sqrt x }}{{x + \sqrt x }}\)

\(P = \left( {\frac{{\sqrt x + 1 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right):\frac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\)

\(P = \frac{{\sqrt x + 1 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }}\)

\(P = \frac{{\sqrt x + 1 + x}}{{\sqrt x }}\)

b) Thay x = 4 vào ta có: \(P = \frac{{\sqrt 4 + 1 + 4}}{{\sqrt 4 }} = \frac{{2 + 1 + 4}}{2} = \frac{7}{2}\)

c) Khi \(P = \frac{{13}}{3}\) thì \(\frac{{13}}{3} = \frac{{\sqrt x + 1 + x}}{{\sqrt x }}\)

⇔ \(3x + 3\sqrt x + 3 = 13\sqrt x \)

⇔ \(3x - 10\sqrt x + 3 = 0\)

⇔ \(3x - 9\sqrt x - \sqrt x + 3 = 0\)

⇔ \(\sqrt x \left( {3\sqrt x - 1} \right) - 3\left( {3\sqrt x - 1} \right) = 0\)

⇔ \(\left( {\sqrt x - 3} \right)\left( {3\sqrt x - 1} \right) = 0\)

⇔ \(\left[ \begin{array}{l}\sqrt x - 3 = 0\\3\sqrt x - 1 = 0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x = 9\\x = \frac{1}{9}\end{array} \right.\).