Đáp án:

$\begin{array}{l}
1)a){x^2} - 4 = 0\\
 \Rightarrow {x^2} = 4\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2
\end{array} \right.\\
b){x^2} + 9 = 0\\
 \Rightarrow {x^2} =  - 9\left( {ktm} \right)\\
Vay\,pt\,vo\,nghiem\\
c)\left( {x - 3} \right).\left( {2x + 7} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x - 3 = 0\\
2x + 7 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 3\\
x =  - \frac{7}{2}
\end{array} \right.\\
d)\left| x \right| + x = 0\\
 \Rightarrow \left| x \right| =  - x\\
 \Rightarrow x \le 0\\
e)\left| x \right| - x = 0\\
 \Rightarrow \left| x \right| = x\\
 \Rightarrow x \ge 0\\
2)a){\left( {x - 3,5} \right)^2} + 1\\
Do:{\left( {x - 3,5} \right)^2} \ge 0\forall x\\
 \Rightarrow {\left( {x - 3,5} \right)^2} + 1 \ge 1\forall x\\
 \Rightarrow GTNN:1 \Leftrightarrow x = 3,5\\
b){\left( {2x - 3} \right)^4} - 2 \ge  - 2\\
 \Rightarrow GTNN =  - 2 \Leftrightarrow x = \frac{3}{2}\\
3)a)\sqrt 2  - {x^2}\\
Do:{x^2} \ge 0\\
 \Rightarrow  - {x^2} \le 0\\
 \Rightarrow \sqrt 2  - {x^2} \le \sqrt 2 \\
 \Rightarrow GTLN = \sqrt 2  \Leftrightarrow x = 0\\
b) - {\left( {x - \sqrt 3 } \right)^2} + 1 \le 1\\
 \Rightarrow GTLN = 1 \Leftrightarrow x = \sqrt 3 \\
4)P\left( 1 \right) = 100 + 99 + 98 + ... + 2 + 1\\
 = \frac{{\left( {100 + 1} \right).100}}{2}\\
 = 5050\\
5)\\
P\left( x \right) = {x^{99}} - 100{x^{98}} + 100{x^{97}} - 100{x^{96}} + \\
 + ... + 100x - 1\\
 = {x^{99}} - 100{x^{97}}\left( {x - 1} \right) - 100.{x^{95}}\left( {x - 1} \right) - ...\\
 - 100x\left( {x - 1} \right) - 1\\
 = \left( {{x^{99}} - 1} \right) + \left( {x - 1} \right).\left( { - 100{x^{97}} - 100{x^{95}} - ... - 100x} \right)\\
 = \left( {x - 1} \right)\left( {{x^{98}} + {x^{97}} + ... + x + 1 - 100{x^{97}} - 100{x^{95}} - .. - 100x} \right)\\
 = \left( {x - 1} \right).\left( {{x^{98}} - 99{x^{97}} + {x^{96}} - 99{x^{95}} + ... + {x^2} - 99x} \right)\\
 = \left( {x - 1} \right)\left( {x - 99} \right)\left( {{x^{97}} + {x^{95}} + .. + x} \right)\\
 \Rightarrow P\left( {99} \right) = \left( {99 - 1} \right).\left( {99 - 99} \right).\left( {{{99}^{97}} + ... + 99} \right) = 0
\end{array}$