Đáp án:
Giải thích các bước giải:
Câu 1:
\(\begin{array}{l}
a.ĐK:x \ge 0;x \ne 1\\
A = \left( {\frac{{x - \sqrt x + \sqrt x - 1}}{{\sqrt x - 1}}} \right):\left( {\frac{{x + 2\sqrt x - \sqrt x - 2}}{{\sqrt x + 2}}} \right)\\
= \frac{{(\sqrt x - 1).(\sqrt x + 1)}}{{\sqrt x - 1}}.\frac{{\sqrt x + 2}}{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 2} \right)}}\\
= \frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b.A(9) = \frac{{\sqrt 9 + 1}}{{\sqrt 9 - 1}} = 2\\
c.A = 5 = \frac{{\sqrt x + 1}}{{\sqrt x - 1}} \to \sqrt x + 1 = 5\sqrt x - 5 \to 4\sqrt x = 6\\
\to x = \frac{9}{4}(TM)\\
d.A < 1 \to \frac{{\sqrt x + 1}}{{\sqrt x - 1}} < 1 \to \frac{{\sqrt x + 1 - \sqrt x + 1}}{{\sqrt x - 1}} < 0\\
\to \frac{2}{{\sqrt x - 1}} < 0 \Leftrightarrow \sqrt x - 1 < 0\\
\Leftrightarrow x = 0(TM)
\end{array}\)
e. A nguyên
\( \Leftrightarrow A = \frac{{\sqrt x + 1}}{{\sqrt x - 1}} = \frac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}} = 1 + \frac{2}{{\sqrt x - 1}}\) nguyên
⇔\({\sqrt x - 1}\) là Ư(2)
\( \Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 1 = 2\\
\sqrt x - 1 = - 2(l)\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 9(TM)\\
x = 4(TM)\\
x = 0(TM)
\end{array} \right.\)
