Đáp án:
$\begin{array}{l}
a){x^{2018}} \ge 0;{x^{2020}} \ge 0\\
\Rightarrow 4{x^{2018}} + 6{x^{2020}} \ge 0\\
\Rightarrow 4{x^{2018}} + 6{x^{2020}} + 2020 \ge 2020 > 0
\end{array}$
Vậy đa thức ko có nghiệm trong R.
$\begin{array}{l}
b)A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^4}}} + \dfrac{1}{{{2^6}}} + ... + \dfrac{1}{{{2^{100}}}}\\
\Rightarrow {2^2}.A = 1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{98}}}}\\
\Rightarrow 4.A - A = \left( {1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{98}}}}} \right)\\
- \left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{98}}}} + \dfrac{1}{{{2^{100}}}}} \right)\\
\Rightarrow 3.A = 1 - \dfrac{1}{{{2^{100}}}}\\
\Rightarrow A = \dfrac{1}{3} - \dfrac{1}{{{{3.2}^{100}}}} < \dfrac{1}{3}\\
Vậy\,A < \dfrac{1}{3}
\end{array}$
