Đáp án:

\(\begin{array}{l}
1,\\
a,\\
A = 5\\
b,\\
B = 5\sqrt 3  + 1\\
2,\\
a,\\
x = 79\\
b,\\
x = \dfrac{3}{8}\\
3,\\
a,\\
P = \dfrac{3}{{\sqrt x  + 3}}\\
b,\\
P = \dfrac{{3\sqrt {11} }}{{11}}\\
c,\\
0 \le x < 9\\
d,\\
x = 0
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
A = \dfrac{{\sqrt 7  + \sqrt 3 }}{{\sqrt 7  - \sqrt 3 }} + \dfrac{{\sqrt 7  - \sqrt 3 }}{{\sqrt 7  + \sqrt 3 }}\\
 = \dfrac{{{{\left( {\sqrt 7  + \sqrt 3 } \right)}^2} + {{\left( {\sqrt 7  - \sqrt 3 } \right)}^2}}}{{\left( {\sqrt 7  - \sqrt 3 } \right)\left( {\sqrt 7  + \sqrt 3 } \right)}}\\
 = \dfrac{{\left( {{{\sqrt 7 }^2} + 2.\sqrt 7 .\sqrt 3  + {{\sqrt 3 }^2}} \right) + \left( {{{\sqrt 7 }^2} - 2.\sqrt 7 .\sqrt 3  + {{\sqrt 3 }^2}} \right)}}{{{{\sqrt 7 }^2} - {{\sqrt 3 }^2}}}\\
 = \dfrac{{\left( {7 + 2\sqrt {21}  + 3} \right) + \left( {7 - 2\sqrt {21}  + 3} \right)}}{{7 - 3}}\\
 = \dfrac{{10 + 2\sqrt {21}  + 10 - 2\sqrt {21} }}{4}\\
 = \dfrac{{20}}{4} = 5\\
b,\\
B = 2\sqrt {27}  + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}}  - \dfrac{4}{{\sqrt 3  + 1}}\\
 = 2.\sqrt {9.3}  + \left| {1 - \sqrt 3 } \right| - \dfrac{{4.\left( {\sqrt 3  - 1} \right)}}{{\left( {\sqrt 3  + 1} \right)\left( {\sqrt 3  - 1} \right)}}\\
 = 2.\sqrt {{3^2}.3}  + \left( {\sqrt 3  - 1} \right) - \dfrac{{4.\left( {\sqrt 3  - 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}}\\
 = 2.3\sqrt 3  + \sqrt 3  - 1 - \dfrac{{4.\left( {\sqrt 3  - 1} \right)}}{2}\\
 = 6\sqrt 3  + \sqrt 3  - 1 - 2\left( {\sqrt 3  - 1} \right)\\
 = 7\sqrt 3  - 1 - 2\sqrt 3  + 2\\
 = 5\sqrt 3  + 1\\
2,\\
a,\\
DKXD:\,\,\,x \ge  - 2\\
\dfrac{1}{5}\sqrt {25x + 50}  - 5\sqrt {x + 2}  + \sqrt {9x + 18}  + 9 = 0\\
 \Leftrightarrow \dfrac{1}{5}.\sqrt {25.\left( {x + 2} \right)}  - 5\sqrt {x + 2}  + \sqrt {9.\left( {x + 2} \right)}  + 9 = 0\\
 \Leftrightarrow \dfrac{1}{5}.\sqrt {{5^2}\left( {x + 2} \right)}  - 5\sqrt {x + 2}  + \sqrt {{3^2}.\left( {x + 2} \right)}  + 9 = 0\\
 \Leftrightarrow \dfrac{1}{5}.5\sqrt {x + 2}  - 5\sqrt {x + 2}  + 3\sqrt {x + 2}  + 9 = 0\\
 \Leftrightarrow \sqrt {x + 2}  - 5\sqrt {x + 2}  + 3\sqrt {x + 2}  + 9 = 0\\
 \Leftrightarrow  - \sqrt {x + 2}  + 9 = 0\\
 \Leftrightarrow \sqrt {x + 2}  = 9\\
 \Leftrightarrow x + 2 = {9^2}\\
 \Leftrightarrow x + 2 = 81\\
 \Leftrightarrow x = 79\\
b,\\
DKXD:\,\,\,{x^2} - 4x + 4 \ge 0 \Leftrightarrow {\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall \,x\\
\sqrt {{x^2} - 4x + 4}  = 7x - 1\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.2 + {2^2}}  = 7x - 1\\
 \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}}  = 7x - 1\\
 \Leftrightarrow \left| {x - 2} \right| = 7x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
7x - 1 \ge 0\\
\left[ \begin{array}{l}
x - 2 = 7x - 1\\
x - 2 =  - 7x + 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{7}\\
\left[ \begin{array}{l}
x - 2 - 7x + 1 = 0\\
x - 2 + 7x - 1 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{7}\\
\left[ \begin{array}{l}
 - 6x - 1 = 0\\
8x - 3 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{7}\\
\left[ \begin{array}{l}
x =  - \dfrac{1}{6}\\
x = \dfrac{3}{8}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{3}{8}\\
3,\\
a,\\
P = \left( {\dfrac{{3x + 3}}{{x - 9}} - \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{3 - \sqrt x }}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\
 = \left( {\dfrac{{3x + 3}}{{{{\sqrt x }^2} - {3^2}}} - \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} - \dfrac{{\sqrt x }}{{\sqrt x  - 3}}} \right):\dfrac{{2\sqrt x  - 2 - \left( {\sqrt x  - 3} \right)}}{{\sqrt x  - 3}}\\
 = \left( {\dfrac{{3x + 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} - \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} - \dfrac{{\sqrt x }}{{\sqrt x  - 3}}} \right):\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\
 = \dfrac{{3x + 3 - 2\sqrt x .\left( {\sqrt x  - 3} \right) - \sqrt x .\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3x + 3 - \left( {2x - 6\sqrt x } \right) - \left( {x + 3\sqrt x } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3x + 3 - 2x + 6\sqrt x  - x - 3\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3\sqrt x  + 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3.\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\
 = \dfrac{3}{{\sqrt x  + 3}}\\
b,\\
x = 20 - 6\sqrt {11}  = 11 - 6\sqrt {11}  + 9\\
 = {\sqrt {11} ^2} - 2.\sqrt {11} .3 + {3^2} = {\left( {\sqrt {11}  - 3} \right)^2}\\
 \Rightarrow \sqrt x  = \sqrt {{{\left( {\sqrt {11}  - 3} \right)}^2}}  = \left| {\sqrt {11}  - 3} \right| = \sqrt {11}  - 3\\
 \Rightarrow P = \dfrac{3}{{\left( {\sqrt {11}  - 3} \right) + 3}} = \dfrac{3}{{\sqrt {11} }} = \dfrac{{3\sqrt {11} }}{{11}}\\
c,\\
P > \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} > \dfrac{1}{2}\\
 \Leftrightarrow 3.2 > 1.\left( {\sqrt x  + 3} \right)\\
 \Leftrightarrow 6 > \sqrt x  + 3\\
 \Leftrightarrow \sqrt x  < 3\\
 \Leftrightarrow 0 \le x < 9\\
d,\\
Q = \dfrac{{2P\sqrt x }}{3} = \dfrac{{2.\dfrac{3}{{\sqrt x  + 3}}.\sqrt x }}{3} = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {2\sqrt x  + 6} \right) - 6}}{{\sqrt x  + 3}} = \dfrac{{2.\left( {\sqrt x  + 3} \right) - 6}}{{\sqrt x  + 3}} = 2 - \dfrac{6}{{\sqrt x  + 3}}\\
Q \in Z \Rightarrow \dfrac{6}{{\sqrt x  + 3}} \in Z\\
x \in Z \Rightarrow \left( {\sqrt x  + 3} \right) \in Ư\left( 6 \right)\\
\sqrt x  + 3 \ge 3,\,\,\,\forall x \ge 0,x \ne 9\\
 \Rightarrow \sqrt x  + 3 \in \left\{ {3;6} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ {0;3} \right\}\\
 \Rightarrow x \in \left\{ {0;9} \right\}\\
x \ne 9 \Rightarrow x = 0
\end{array}\)

\(\begin{array}{l}
1,\\
a,\\
A = \dfrac{{\sqrt 7  + \sqrt 3 }}{{\sqrt 7  - \sqrt 3 }} + \dfrac{{\sqrt 7  - \sqrt 3 }}{{\sqrt 7  + \sqrt 3 }}\\
 = \dfrac{{{{\left( {\sqrt 7  + \sqrt 3 } \right)}^2} + {{\left( {\sqrt 7  - \sqrt 3 } \right)}^2}}}{{\left( {\sqrt 7  - \sqrt 3 } \right)\left( {\sqrt 7  + \sqrt 3 } \right)}}\\
 = \dfrac{{\left( {{{\sqrt 7 }^2} + 2.\sqrt 7 .\sqrt 3  + {{\sqrt 3 }^2}} \right) + \left( {{{\sqrt 7 }^2} - 2.\sqrt 7 .\sqrt 3  + {{\sqrt 3 }^2}} \right)}}{{{{\sqrt 7 }^2} - {{\sqrt 3 }^2}}}\\
 = \dfrac{{\left( {7 + 2\sqrt {21}  + 3} \right) + \left( {7 - 2\sqrt {21}  + 3} \right)}}{{7 - 3}}\\
 = \dfrac{{10 + 2\sqrt {21}  + 10 - 2\sqrt {21} }}{4}\\
 = \dfrac{{20}}{4} = 5\\
b,\\
B = 2\sqrt {27}  + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}}  - \dfrac{4}{{\sqrt 3  + 1}}\\
 = 2.\sqrt {9.3}  + \left| {1 - \sqrt 3 } \right| - \dfrac{{4.\left( {\sqrt 3  - 1} \right)}}{{\left( {\sqrt 3  + 1} \right)\left( {\sqrt 3  - 1} \right)}}\\
 = 2.\sqrt {{3^2}.3}  + \left( {\sqrt 3  - 1} \right) - \dfrac{{4.\left( {\sqrt 3  - 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}}\\
 = 2.3\sqrt 3  + \sqrt 3  - 1 - \dfrac{{4.\left( {\sqrt 3  - 1} \right)}}{2}\\
 = 6\sqrt 3  + \sqrt 3  - 1 - 2\left( {\sqrt 3  - 1} \right)\\
 = 7\sqrt 3  - 1 - 2\sqrt 3  + 2\\
 = 5\sqrt 3  + 1\\
2,\\
a,\\
DKXD:\,\,\,x \ge  - 2\\
\dfrac{1}{5}\sqrt {25x + 50}  - 5\sqrt {x + 2}  + \sqrt {9x + 18}  + 9 = 0\\
 \Leftrightarrow \dfrac{1}{5}.\sqrt {25.\left( {x + 2} \right)}  - 5\sqrt {x + 2}  + \sqrt {9.\left( {x + 2} \right)}  + 9 = 0\\
 \Leftrightarrow \dfrac{1}{5}.\sqrt {{5^2}\left( {x + 2} \right)}  - 5\sqrt {x + 2}  + \sqrt {{3^2}.\left( {x + 2} \right)}  + 9 = 0\\
 \Leftrightarrow \dfrac{1}{5}.5\sqrt {x + 2}  - 5\sqrt {x + 2}  + 3\sqrt {x + 2}  + 9 = 0\\
 \Leftrightarrow \sqrt {x + 2}  - 5\sqrt {x + 2}  + 3\sqrt {x + 2}  + 9 = 0\\
 \Leftrightarrow  - \sqrt {x + 2}  + 9 = 0\\
 \Leftrightarrow \sqrt {x + 2}  = 9\\
 \Leftrightarrow x + 2 = {9^2}\\
 \Leftrightarrow x + 2 = 81\\
 \Leftrightarrow x = 79\\
b,\\
DKXD:\,\,\,{x^2} - 4x + 4 \ge 0 \Leftrightarrow {\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall \,x\\
\sqrt {{x^2} - 4x + 4}  = 7x - 1\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.2 + {2^2}}  = 7x - 1\\
 \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}}  = 7x - 1\\
 \Leftrightarrow \left| {x - 2} \right| = 7x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
7x - 1 \ge 0\\
\left[ \begin{array}{l}
x - 2 = 7x - 1\\
x - 2 =  - 7x + 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{7}\\
\left[ \begin{array}{l}
x - 2 - 7x + 1 = 0\\
x - 2 + 7x - 1 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{7}\\
\left[ \begin{array}{l}
 - 6x - 1 = 0\\
8x - 3 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{7}\\
\left[ \begin{array}{l}
x =  - \dfrac{1}{6}\\
x = \dfrac{3}{8}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{3}{8}\\
3,\\
a,\\
P = \left( {\dfrac{{3x + 3}}{{x - 9}} - \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{3 - \sqrt x }}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\
 = \left( {\dfrac{{3x + 3}}{{{{\sqrt x }^2} - {3^2}}} - \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} - \dfrac{{\sqrt x }}{{\sqrt x  - 3}}} \right):\dfrac{{2\sqrt x  - 2 - \left( {\sqrt x  - 3} \right)}}{{\sqrt x  - 3}}\\
 = \left( {\dfrac{{3x + 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} - \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} - \dfrac{{\sqrt x }}{{\sqrt x  - 3}}} \right):\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\
 = \dfrac{{3x + 3 - 2\sqrt x .\left( {\sqrt x  - 3} \right) - \sqrt x .\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3x + 3 - \left( {2x - 6\sqrt x } \right) - \left( {x + 3\sqrt x } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3x + 3 - 2x + 6\sqrt x  - x - 3\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3\sqrt x  + 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
 = \dfrac{{3.\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\
 = \dfrac{3}{{\sqrt x  + 3}}\\
b,\\
x = 20 - 6\sqrt {11}  = 11 - 6\sqrt {11}  + 9\\
 = {\sqrt {11} ^2} - 2.\sqrt {11} .3 + {3^2} = {\left( {\sqrt {11}  - 3} \right)^2}\\
 \Rightarrow \sqrt x  = \sqrt {{{\left( {\sqrt {11}  - 3} \right)}^2}}  = \left| {\sqrt {11}  - 3} \right| = \sqrt {11}  - 3\\
 \Rightarrow P = \dfrac{3}{{\left( {\sqrt {11}  - 3} \right) + 3}} = \dfrac{3}{{\sqrt {11} }} = \dfrac{{3\sqrt {11} }}{{11}}\\
c,\\
P > \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} > \dfrac{1}{2}\\
 \Leftrightarrow 3.2 > 1.\left( {\sqrt x  + 3} \right)\\
 \Leftrightarrow 6 > \sqrt x  + 3\\
 \Leftrightarrow \sqrt x  < 3\\
 \Leftrightarrow 0 \le x < 9\\
d,\\
Q = \dfrac{{2P\sqrt x }}{3} = \dfrac{{2.\dfrac{3}{{\sqrt x  + 3}}.\sqrt x }}{3} = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {2\sqrt x  + 6} \right) - 6}}{{\sqrt x  + 3}} = \dfrac{{2.\left( {\sqrt x  + 3} \right) - 6}}{{\sqrt x  + 3}} = 2 - \dfrac{6}{{\sqrt x  + 3}}\\
Q \in Z \Rightarrow \dfrac{6}{{\sqrt x  + 3}} \in Z\\
x \in Z \Rightarrow \left( {\sqrt x  + 3} \right) \in Ư\left( 6 \right)\\
\sqrt x  + 3 \ge 3,\,\,\,\forall x \ge 0,x \ne 9\\
 \Rightarrow \sqrt x  + 3 \in \left\{ {3;6} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ {0;3} \right\}\\
 \Rightarrow x \in \left\{ {0;9} \right\}\\
x \ne 9 \Rightarrow x = 0
\end{array}\)