Đáp án:

\(\begin{array}{l}
a)M = \dfrac{1}{{\sqrt x }}\\
b)2 - \sqrt 3 \\
c)x = 9
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)M = \left[ {\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}} + \dfrac{{\sqrt x }}{{\sqrt x  + 1}} - \dfrac{{x + 2\sqrt x  - 3}}{{x - 1}}} \right].\dfrac{{x - 1}}{{\sqrt x \left( {x - \sqrt x  + 4} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2} + \sqrt x \left( {\sqrt x  - 1} \right) - x - 2\sqrt x  + 3}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x \left( {x - \sqrt x  + 4} \right)}}\\
 = \dfrac{{x + 2\sqrt x  + 1 + x - \sqrt x  - x - 2\sqrt x  + 3}}{{\sqrt x \left( {x - \sqrt x  + 4} \right)}}\\
 = \dfrac{{x - \sqrt x  + 4}}{{\sqrt x \left( {x - \sqrt x  + 4} \right)}}\\
 = \dfrac{1}{{\sqrt x }}\\
b)Thay:x = 7 + 4\sqrt 3 \\
 = 4 + 2.2.\sqrt 3  + 3\\
 = {\left( {2 + \sqrt 3 } \right)^2}\\
 \to M = \dfrac{1}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} }} = \dfrac{1}{{2 + \sqrt 3 }}\\
 = 2 - \sqrt 3 \\
c)\left( {x - \sqrt x  - 3} \right).M = 1\\
 \to \left( {x - \sqrt x  - 3} \right).\dfrac{1}{{\sqrt x }} = 1\\
 \to x - \sqrt x  - 3 = \sqrt x \\
 \to x - 2\sqrt x  - 3 = 0\\
 \to \left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right) = 0\\
 \to \sqrt x  - 3 = 0\left( {do:\sqrt x  + 1 > 0\forall x} \right)\\
 \to x = 9
\end{array}\)