Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
x = 7 - 4\sqrt 3  = 4 - 2.2.\sqrt 3  + 3 = {\left( {2 - \sqrt 3 } \right)^2}\\
 \Rightarrow \sqrt x  = 2 - \sqrt 3 \\
B = \dfrac{2}{{\sqrt x  - 2}} = \dfrac{2}{{\left( {2 - \sqrt 3 } \right) - 2}} = \dfrac{2}{{ - \sqrt 3 }} = \dfrac{{ - 2\sqrt 3 }}{3}\\
b,\\
A = \dfrac{{\sqrt x }}{{x - 4}} + \dfrac{1}{{\sqrt x  - 2}}\\
 = \dfrac{{\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{1}{{\sqrt x  - 2}}\\
 = \dfrac{{\sqrt x  + \left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{2\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
P = \dfrac{B}{A} = \dfrac{2}{{\sqrt x  - 2}}:\dfrac{{2\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{2}{{\sqrt x  - 2}}.\dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}{{2.\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
c,\\
P = \dfrac{4}{3} \Leftrightarrow \dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}} = \dfrac{4}{3}\\
 \Leftrightarrow 3.\left( {\sqrt x  + 2} \right) = 4.\left( {\sqrt x  + 1} \right)\\
 \Leftrightarrow 3\sqrt x  + 6 = 4\sqrt x  + 4\\
 \Leftrightarrow \sqrt x  = 2\\
 \Leftrightarrow x = 4\,\,\,\,\left( {L,\,\,\,x \ne 4} \right)\\
 \Rightarrow ptvn\\
d,\\
\left( {\sqrt x  + 1} \right)P - \sqrt x  - 4\sqrt {x - 1}  + 26 =  - 6x + 10\sqrt {5x} \,\,\,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
 \Leftrightarrow \left( {\sqrt x  + 1} \right).\dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}} - \sqrt x  - 4\sqrt {x - 1}  + 26 =  - 6x + 10\sqrt {5x} \\
 \Leftrightarrow \left( {\sqrt x  + 2} \right) - \sqrt x  - 4\sqrt {x - 1}  + 26 =  - 6x + 10\sqrt {5x} \\
 \Leftrightarrow 2 - 4\sqrt {x - 1}  + 26 + 6x - 10\sqrt {5x}  = 0\\
 \Leftrightarrow 6x - 10\sqrt {5x}  - 4\sqrt {x - 1}  + 28 = 0\\
 \Leftrightarrow \left( {5x - 10\sqrt {5x}  + 25} \right) + \left[ {\left( {x - 1} \right) - 4\sqrt {x - 1}  + 4} \right] = 0\\
 \Leftrightarrow {\left( {\sqrt {5x}  - 5} \right)^2} + {\left( {\sqrt {x - 1}  - 2} \right)^2} = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {\sqrt {5x}  - 5} \right)^2} = 0\\
{\left( {\sqrt {x - 1}  - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {5x}  = 5\\
\sqrt {x - 1}  = 2
\end{array} \right. \Leftrightarrow x = 5
\end{array}\)