Đáp án:

B2:

\(B = \dfrac{{3\sqrt x }}{{\sqrt x  + 3}}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
1)a)\dfrac{{5\left( {\sqrt 5  + 1} \right) - 5\left( {\sqrt 5  - 1} \right)}}{{5 - 1}}\\
 = \dfrac{{5\sqrt 5  + 5 - 5\sqrt 5  + 5}}{4}\\
 = \dfrac{{10}}{4} = \dfrac{5}{2}\\
b)\left| {\sqrt 5  - 3} \right| - \dfrac{{\sqrt 5 }}{5}\\
 = 3 - \sqrt 5  - \dfrac{{\sqrt 5 }}{5} = 3 - \dfrac{{6\sqrt 5 }}{5}\\
2)a)DK:x \ge 1\\
\sqrt {x - 1}  + \sqrt {9\left( {x - 1} \right)}  + \sqrt {4\left( {x - 1} \right)}  = 12\\
 \to \left( {1 + 3 + 2} \right)\sqrt {x - 1}  = 12\\
 \to 6\sqrt {x - 1}  = 12\\
 \to \sqrt {x - 1}  = 2\\
 \to x - 1 = 4\\
 \to x = 5\left( {TM} \right)\\
b)DK:x \ge 5\\
\sqrt {x\left( {x - 5} \right)}  - \sqrt {x - 5}  = 0\\
 \to \sqrt {x - 5} \left( {\sqrt x  - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x - 5 = 0\\
\sqrt x  - 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\left( {TM} \right)\\
x = 1\left( l \right)
\end{array} \right.\\
B2:\\
b)B = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right) - 7\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2x - 6\sqrt x  + x + 4\sqrt x  + 3 - 7\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3x - 9\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3\sqrt x }}{{\sqrt x  + 3}}\\
 \to dpcm
\end{array}\)

\(\begin{array}{l}
c)P = A.B = \dfrac{{x + 7}}{{3\sqrt x }}.\dfrac{{3\sqrt x }}{{\sqrt x  + 3}}\\
 = \dfrac{{x + 7}}{{\sqrt x  + 3}} = \dfrac{{x - 9 + 16}}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) + 16}}{{\sqrt x  + 3}}\\
 = \sqrt x  - 3 + \dfrac{{16}}{{\sqrt x  + 3}}\\
 = \left( {\sqrt x  + 3} \right) + \dfrac{{16}}{{\sqrt x  + 3}} - 6\\
Do:x > 0\\
BDT:Co - si:\left( {\sqrt x  + 3} \right) + \dfrac{{16}}{{\sqrt x  + 3}} \ge 2\sqrt {\left( {\sqrt x  + 3} \right).\dfrac{{16}}{{\sqrt x  + 3}}} \\
 \to \left( {\sqrt x  + 3} \right) + \dfrac{{16}}{{\sqrt x  + 3}} \ge 8\\
 \to \left( {\sqrt x  + 3} \right) + \dfrac{{16}}{{\sqrt x  + 3}} - 6 \ge 2\\
 \to Min = 2\\
 \Leftrightarrow \left( {\sqrt x  + 3} \right) = \dfrac{{16}}{{\sqrt x  + 3}}\\
 \to \left[ \begin{array}{l}
\sqrt x  + 3 = 4\\
\sqrt x  + 3 =  - 4\left( l \right)
\end{array} \right.\\
 \to x = 1
\end{array}\)