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