Đáp án:
\(\begin{array}{l}
a)x > 1\\
b)x > 1;x \ne 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left\{ \begin{array}{l}
x - 1 \ge 0\\
1 - \sqrt x \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 1\\
x \ne 1
\end{array} \right.\\
\to x > 1\\
b)A = \dfrac{1}{{\sqrt x + \sqrt {x - 1} }} - \dfrac{1}{{\sqrt x - \sqrt {x - 1} }} - \dfrac{{x\left( {\sqrt x - 1} \right)}}{{ - \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - \sqrt {x - 1} - \sqrt x - \sqrt {x - 1} }}{{x - \left( {x - 1} \right)}} + x\\
= \dfrac{{ - 2\sqrt {x - 1} }}{1} + x\\
= x - 2\sqrt {x - 1} \\
A > 0\\
\to x - 2\sqrt {x - 1} > 0\\
\to x > 2\sqrt {x - 1} \\
\to {x^2} > 4\left( {x - 1} \right)\\
\to {x^2} - 4x + 4 > 0\\
\to {\left( {x - 2} \right)^2} > 0\\
\to x - 2 \ne 0\\
\to x \ne 2\\
KL:x > 1;x \ne 2
\end{array}\)
