Giải thích các bước giải:
Ta có:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x > 0\\
x \ne \frac{9}{4}\\
x \ne 6
\end{array} \right.\)
\(\begin{array}{l}
A = \frac{{2x\sqrt x + 2\sqrt x - 3x - 3}}{{2\sqrt x - 3}}\\
= \frac{{2\sqrt x \left( {x + 1} \right) - 3\left( {x + 1} \right)}}{{2\sqrt x - 3}}\\
= \frac{{\left( {x + 1} \right)\left( {2\sqrt x - 3} \right)}}{{2\sqrt x - 3}}\\
= x + 1\\
B = \frac{{2{x^2} - 2x}}{{x\sqrt x - 6\sqrt x }} = \frac{{2x\left( {x - 1} \right)}}{{\sqrt x \left( {x - 6} \right)}} = \frac{{2\sqrt x \left( {x - 1} \right)}}{{x - 6}}\\
b,\\
x = \sqrt 4 + \sqrt 7 \Rightarrow A = \sqrt 4 + \sqrt 7 + 1
\end{array}\)