Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
a,\\
B = \left( {\frac{{x + \sqrt x - 1}}{{x\sqrt x - 1}} - \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}} \right):\frac{1}{{\sqrt x - 1}}\\
= \left( {\frac{{x + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}} \right):\frac{1}{{\sqrt x - 1}}\\
= \frac{{x + \sqrt x - 1 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\frac{1}{{\sqrt x - 1}}\\
= \frac{{x + \sqrt x - 1 - \left( {x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\frac{1}{{\sqrt x - 1}}\\
= \frac{{\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {\sqrt x - 1} \right)\\
= \frac{{\sqrt x }}{{x + \sqrt x + 1}}\\
b,\\
B - \frac{1}{3} = \frac{{\sqrt x }}{{x + \sqrt x + 1}} - \frac{1}{3} = \frac{{3\sqrt x - \left( {x + \sqrt x + 1} \right)}}{{3.\left( {x + \sqrt x + 1} \right)}}\\
= \frac{{ - \left( {x - 2\sqrt x + 1} \right)}}{{3.\left( {x + \sqrt x + 1} \right)}} = \frac{{ - {{\left( {\sqrt x - 1} \right)}^2}}}{{3.\left( {x + \sqrt x + 1} \right)}} < 0,\,\,\,\forall x \ge 0,x \ne 1\\
\Rightarrow B < \frac{1}{3}\\
c,\\
B = \frac{1}{{2\sqrt x + 1}}\\
\Leftrightarrow \frac{{\sqrt x }}{{x + \sqrt x + 1}} = \frac{1}{{2\sqrt x + 1}}\\
\Leftrightarrow \sqrt x \left( {2\sqrt x + 1} \right) = x + \sqrt x + 1\\
\Leftrightarrow 2x + \sqrt x = x + \sqrt x + 1\\
\Leftrightarrow x = 1
\end{array}\)
Mà \(x = 1\) không thỏa mãn ĐKXĐ nên phương trình trên vô nghiệm.
