Đáp án:

\(\begin{array}{l}
B1:\\
1)B = 3\\
2)P = \dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\
3)Max = 5\\
B2:\\
1)A = \dfrac{4}{3}\\
2)\dfrac{{3\sqrt x }}{{\sqrt x  - 3}}\\
3)x = 0
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
1)Thay:x = 4\\
 \to B = \dfrac{3}{{\sqrt 4  - 1}} = \dfrac{3}{{2 - 1}} = 3\\
2)A = \dfrac{{6 + \sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - \sqrt x  + 6}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
P = A - B = \dfrac{{x - \sqrt x  + 6}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} - \dfrac{3}{{\sqrt x  - 1}}\\
 = \dfrac{{x - \sqrt x  + 6 - 3\sqrt x  - 3}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - 4\sqrt x  + 3}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\
3)\dfrac{1}{P} = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} = \dfrac{{\sqrt x  - 3 + 4}}{{\sqrt x  - 3}}\\
 = 1 + \dfrac{4}{{\sqrt x  - 3}}\\
\dfrac{1}{P}\max  \Leftrightarrow \dfrac{4}{{\sqrt x  - 3}}\max \\
 \Leftrightarrow \left( {\sqrt x  - 3} \right)\min \\
 \Leftrightarrow \sqrt x  - 3 = 1\\
 \to \sqrt x  = 4\\
 \to x = 16\\
 \to Max = 1 + \dfrac{4}{{\sqrt {16}  - 3}} = 5\\
B2:\\
1)Thay:x = 36\\
 \to A = \dfrac{{2\sqrt {36} }}{{\sqrt {36}  + 3}} = \dfrac{{2.6}}{{6 + 3}} = \dfrac{4}{3}\\
2)B = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right) + 11\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x + 4\sqrt x  + 3 + 11\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x + 15\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
M = A + B = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{x + 15\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + x + 15\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{2x - 6\sqrt x  + x + 15\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3x + 9\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x }}{{\sqrt x  - 3}}\\
3)M = {M^4}\\
 \to {M^4} - M = 0\\
 \to M\left( {{M^3} - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
M = 0\\
{M^3} = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
M = 0\\
M = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{3\sqrt x }}{{\sqrt x  - 3}} = 0\\
\dfrac{{3\sqrt x }}{{\sqrt x  - 3}} = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 0\\
\dfrac{{3\sqrt x  - \sqrt x  + 3}}{{\sqrt x  - 3}} = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
2\sqrt x  + 3 = 0\left( {KTM} \right)
\end{array} \right.\\
 \to x = 0
\end{array}\)