Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
D = \left[ {4;7} \right]\\
\sqrt {x + 3} - \sqrt {7 - x} > \sqrt {2x - 8} \\
\Leftrightarrow \sqrt {x + 3} > \sqrt {7 - x} + \sqrt {2x - 8} \\
\Leftrightarrow {\sqrt {x + 3} ^2} > {\left( {\sqrt {7 - x} + \sqrt {2x - 8} } \right)^2}\\
\Leftrightarrow x + 3 > 7 - x + 2\sqrt {\left( {7 - x} \right)\left( {2x - 8} \right)} + 2x - 8\\
\Leftrightarrow x + 3 > x - 1 + 2\sqrt {\left( {7 - x} \right)\left( {2x - 8} \right)} \\
\Leftrightarrow \sqrt {\left( {7 - x} \right)\left( {2x - 8} \right)} < 2\\
\Leftrightarrow 14x - 56 - 2{x^2} + 8x < 4\\
\Leftrightarrow - 2{x^2} + 22x - 60 < 0\\
\Leftrightarrow {x^2} - 11x + 30 > 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {x - 6} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 6\\
x < 5
\end{array} \right.\\
\Rightarrow x \in \left[ {4;5} \right) \cup \left( {6;7} \right]\\
x \in Z \Rightarrow x \in \left\{ {4;7} \right\}\\
2,\\
a \in \left( {\dfrac{\pi }{2};\pi } \right) \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
{\sin ^2}a + {\cos ^2}a = 1\\
\cos a < 0 \Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - \dfrac{{2\sqrt 2 }}{3}\\
\tan a = \dfrac{{\sin a}}{{\cos a}} = - \dfrac{1}{{2\sqrt 2 }},\,\,\,\,\cot a = \dfrac{{\cos a}}{{\sin a}} = - 2\sqrt 2 \\
P = \dfrac{{2\tan a + 3\cot a + 1}}{{\tan a + \cot a}} = \dfrac{{26 - 2\sqrt 2 }}{9}\\
\Rightarrow a = 26;\,\,\,b = - 2;\,\,\,c = 9\\
\Rightarrow a + b + c = 33\\
3,\\
{d_{\left( {I,\Delta } \right)}} = \dfrac{{\left| {3.1 + 4.1 + 13} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \dfrac{{20}}{5} = 4\\
\Rightarrow R = \sqrt {{d^2}_{\left( {I,\Delta } \right)} + {{\left( {\dfrac{l}{2}} \right)}^2}} = \sqrt {{4^2} + {{\left( {\dfrac{6}{2}} \right)}^2}} = 5\\
\Rightarrow \left( C \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = {R^2}\\
\Leftrightarrow {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 25\\
4,\\
P = {\left( {\dfrac{{\cos x + \cot x}}{{1 + \sin x}}} \right)^2} + 1 = {\left( {\dfrac{{\cos x + \dfrac{{\cos x}}{{\sin x}}}}{{1 + \sin x}}} \right)^2} + 1\\
= {\left( {\dfrac{{\cos x.\left( {1 + \dfrac{1}{{\sin x}}} \right)}}{{1 + \sin x}}} \right)^2} + 1 = {\left( {\dfrac{{\cos x.\dfrac{{\sin x + 1}}{{\sin x}}}}{{1 + \sin x}}} \right)^2} + 1\\
= {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2} + 1 = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\sin }^2}x}} = \dfrac{1}{{{{\sin }^2}x}}
\end{array}\)
