Đáp án:
b) \(0 \le x < \dfrac{{25}}{9};x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
A = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + \sqrt x - 1 - x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\dfrac{{3 - \sqrt x + \sqrt x - 1}}{{\sqrt x - 1}}\\
= \dfrac{{x + \sqrt x + \sqrt x - x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)\sqrt A < \dfrac{1}{2}\\
\to A < \dfrac{1}{4}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} < \dfrac{1}{4}\\
\to \dfrac{{4\sqrt x - 4 - \sqrt x - 1}}{{4\left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{3\sqrt x - 5}}{{4\left( {\sqrt x + 1} \right)}} < 0\\
\to 3\sqrt x - 5 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to x < \dfrac{{25}}{9}\\
KL:0 \le x < \dfrac{{25}}{9};x \ne 1
\end{array}\)
( sửa chữ P thành chữ A nha b)
