Đáp án:
b) \(\left[ \begin{array}{l}
\dfrac{1}{2} > m;m \ne 0\\
m > 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right).\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \sqrt x - 1\\
b)B = m\sqrt x - 2m\\
\to \sqrt x - 1 = m\sqrt x - 2m\\
\to \left( {1 - m} \right)\sqrt x = 1 - 2m\\
\to \sqrt x = \dfrac{{1 - 2m}}{{1 - m}}
\end{array}\)
Để tồn tại giá trị x TMĐK
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
\dfrac{{1 - 2m}}{{1 - m}} > 0\\
\dfrac{{1 - 2m}}{{1 - m}} \ne 1\\
1 - m \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - 2m > 0\\
1 - m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - 2m < 0\\
1 - m < 0
\end{array} \right.
\end{array} \right.\\
1 - 2m \ne 1 - m\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\dfrac{1}{2} > m\\
m > 1
\end{array} \right.\\
m \ne \left\{ {0;1} \right\}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{1}{2} > m;m \ne 0\\
m > 1
\end{array} \right.
\end{array}\)
