Đáp án:

\(\begin{array}{l}
1,\\
a,\\
x < 3\\
b,\\
\left\{ \begin{array}{l}
x \ge 0\\
y \ge 0\\
x \ne y
\end{array} \right.\\
2,\\
a,\\
A = 1\\
b,\\
B = \sqrt 5 \\
3,\\
a,\\
x = 0\\
b,\\
x = 14\\
4,\\
a,\\
\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
b,\\
P = \dfrac{{ - 3}}{{\sqrt x  - 2}}\\
b,\\
x \in \left\{ {1;9;25} \right\}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
Các\,\,biểu\,\,thức\,\,đã\,\,cho\,\,có\,\,nghĩa\,\,khi:\\
a,\\
\left\{ \begin{array}{l}
\dfrac{{ - 2}}{{x - 3}} \ge 0\\
x - 3 \ne 0
\end{array} \right. \Leftrightarrow x - 3 < 0 \Leftrightarrow x < 3\\
b,\\
\left\{ \begin{array}{l}
x \ge 0\\
y \ge 0\\
\sqrt x  - \sqrt y  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
y \ge 0\\
\sqrt x  \ne \sqrt y 
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
y \ge 0\\
x \ne y
\end{array} \right.\\
2,\\
a,\\
A = \left( {\sqrt {5 - 2\sqrt 6 }  + \sqrt 2 } \right).\dfrac{1}{{\sqrt 3 }}\\
 = \left( {\sqrt {3 - 2\sqrt 6  + 2}  + \sqrt 2 } \right).\dfrac{1}{{\sqrt 3 }}\\
 = \left( {\sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 .\sqrt 2  + {{\sqrt 2 }^2}}  + \sqrt 2 } \right).\dfrac{1}{{\sqrt 3 }}\\
 = \left( {\sqrt {{{\left( {\sqrt 3  - \sqrt 2 } \right)}^2}}  + \sqrt 2 } \right).\dfrac{1}{{\sqrt 3 }}\\
 = \left( {\left| {\sqrt 3  - \sqrt 2 } \right| + \sqrt 2 } \right).\dfrac{1}{{\sqrt 3 }}\\
 = \left( {\sqrt 3  - \sqrt 2  + \sqrt 2 } \right).\dfrac{1}{{\sqrt 3 }}\\
 = \sqrt 3 .\dfrac{1}{{\sqrt 3 }}\\
 = 1\\
b,\\
B = 2 + \sqrt {17 - 4\sqrt {9 + 4\sqrt 5 } } \\
 = 2 + \sqrt {17 - 4\sqrt {5 + 4\sqrt 5  + 4} } \\
 = 2 + \sqrt {17 - 4\sqrt {{{\sqrt 5 }^2} + 2.\sqrt 5 .2 + {2^2}} } \\
 = 2 + \sqrt {17 - 4\sqrt {{{\left( {\sqrt 5  + 2} \right)}^2}} } \\
 = 2 + \sqrt {17 - 4\left| {\sqrt 5  + 2} \right|} \\
 = 2 + \sqrt {17 - 4\left( {\sqrt 5  + 2} \right)} \\
 = 2 + \sqrt {17 - 4\sqrt 5  - 8} \\
 = 2 + \sqrt {9 - 4\sqrt 5 } \\
 = 2 + \sqrt {5 - 4\sqrt 5  + 4} \\
 = 2 + \sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .2 + {2^2}} \\
 = 2 + \sqrt {{{\left( {\sqrt 5  - 2} \right)}^2}} \\
 = 2 + \left| {\sqrt 5  - 2} \right|\\
 = 2 + \sqrt 5  - 2\\
 = \sqrt 5 \\
3,\\
a,\\
DKXD:\,\,\,{x^2} - 2x + 4 \ge 0 \Leftrightarrow {\left( {x - 1} \right)^2} + 3 \ge 0,\,\,\,\,\forall x\\
\sqrt {{x^2} - 2x + 4}  = x + 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 2 \ge 0\\
{x^2} - 2x + 4 = {\left( {x + 2} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 2\\
{x^2} - 2x + 4 = {x^2} + 2.x.2 + {2^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 2\\
{x^2} - 2x + 4 = {x^2} + 4x + 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 2\\
 - 2x = 4x
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 2\\
6x = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 2\\
x = 0
\end{array} \right.\\
 \Leftrightarrow x = 0\\
b,\\
DKXD:\,\,\,x \ge 5\\
\sqrt {4x - 20}  + 3\sqrt {\dfrac{{x - 5}}{9}}  - \dfrac{1}{4}\sqrt {16x - 80}  = 6\\
 \Leftrightarrow \sqrt {4.\left( {x - 5} \right)}  + 3\sqrt {\dfrac{1}{9}.\left( {x - 5} \right)}  - \dfrac{1}{4}\sqrt {16\left( {x - 5} \right)}  = 6\\
 \Leftrightarrow \sqrt {{2^2}.\left( {x - 5} \right)}  + 3.\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}.\left( {x - 5} \right)}  - \dfrac{1}{4}.\sqrt {{4^2}.\left( {x - 5} \right)}  = 6\\
 \Leftrightarrow 2\sqrt {x - 5}  + 3.\dfrac{1}{3}.\sqrt {x - 5}  - \dfrac{1}{4}.4\sqrt {x - 5}  = 6\\
 \Leftrightarrow 2\sqrt {x - 5}  + \sqrt {x - 5}  - \sqrt {x - 5}  = 6\\
 \Leftrightarrow 2\sqrt {x - 5}  = 6\\
 \Leftrightarrow \sqrt {x - 5}  = 3\\
 \Leftrightarrow x - 5 = {3^2}\\
 \Leftrightarrow x - 5 = 9\\
 \Leftrightarrow x = 14\\
4,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - 4 \ne 0\\
\sqrt x  + 2 \ne 0\\
2 - \sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x - 4 \ne 0\\
\sqrt x  \ne 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
b,\\
P = \left( {\dfrac{{\sqrt x }}{{x - 4}} + \dfrac{1}{{\sqrt x  + 2}} + \dfrac{2}{{2 - \sqrt x }}} \right):\left( {\sqrt x  + \dfrac{{6 - x}}{{\sqrt x  + 2}} - 2} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{{{\sqrt x }^2} - {2^2}}} + \dfrac{1}{{\sqrt x  + 2}} - \dfrac{2}{{\sqrt x  - 2}}} \right):\dfrac{{\sqrt x .\left( {\sqrt x  + 2} \right) + 6 - x - 2.\left( {\sqrt x  + 2} \right)}}{{\sqrt x  + 2}}\\
 = \left( {\dfrac{{\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{1}{{\sqrt x  + 2}} - \dfrac{2}{{\sqrt x  - 2}}} \right):\dfrac{{x + 2\sqrt x  + 6 - x - 2\sqrt x  - 4}}{{\sqrt x  + 2}}\\
 = \dfrac{{\sqrt x  + \left( {\sqrt x  - 2} \right) - 2.\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\dfrac{2}{{\sqrt x  + 2}}\\
 = \dfrac{{\sqrt x  + \sqrt x  - 2 - 2\sqrt x  - 4}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\dfrac{2}{{\sqrt x  + 2}}\\
 = \dfrac{{ - 6}}{{\left( {\sqrt x  - 2} \right).\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  + 2}}{2}\\
 = \dfrac{{ - 3}}{{\sqrt x  - 2}}\\
b,\\
P \in Z \Leftrightarrow \dfrac{{ - 3}}{{\sqrt x  - 2}} \in Z \Leftrightarrow \dfrac{3}{{\sqrt x  - 2}} \in Z\\
x \in Z \Rightarrow \left( {\sqrt x  - 2} \right) \in Ư\left( 3 \right) = \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ { - 1;1;3;5} \right\}\\
\sqrt x  \ge 0 \Rightarrow \sqrt x  \in \left\{ {1;3;5} \right\}\\
 \Rightarrow x \in \left\{ {1;9;25} \right\}
\end{array}\)