Đáp án:
b) 2) \(x = \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)Thay:x = 9\\
\to A = \dfrac{{\sqrt 9 + 1}}{{\sqrt 9 - 1}} = \dfrac{4}{2} = 2\\
b)P = \left[ {\dfrac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right].\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
2)2P = 2\sqrt x + 5\\
\to 2.\dfrac{{\sqrt x + 1}}{{\sqrt x }} = 2\sqrt x + 5\\
\to 2\sqrt x + 2 = 2x + 5\sqrt x \\
\to 2x + 3\sqrt x - 2 = 0\\
\to \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) = 0\\
\to 2\sqrt x - 1 = 0\left( {do:\sqrt x + 2 > 0\forall x > 0} \right)\\
\to \sqrt x = \dfrac{1}{2}\\
\to x = \dfrac{1}{4}
\end{array}\)
