Giải thích các bước giải:
B1:
$\begin{array}{l}
a){x^{2018}}{y^{2019}} + 4{x^{2014}}{y^{2015}} = {x^{2014}}{y^{2015}}\left( {{x^4}{y^4} + 4} \right)\\
b){x^2}y + x{y^2} + x + y = 2020\\
\Leftrightarrow xy\left( {x + y} \right) + x + y = 2020\\
\Leftrightarrow \left( {x + y} \right)\left( {xy + 1} \right) = 2020\\
\Leftrightarrow 20\left( {x + y} \right) = 2020\left( {xy = 19} \right)\\
\Leftrightarrow x + y = 101\\
\Rightarrow {x^2} + {y^2} - 63 = {\left( {x + y} \right)^2} - 2xy - 63 = {101^2} - 2.19 - 63 = 10100
\end{array}$
B2:
$\begin{array}{l}
a)P = \dfrac{{2{x^2} + 3x + 3}}{{2x - 1}} = \dfrac{{\left( {2x - 1} \right)\left( {x + 2} \right) + 5}}{{2x - 1}} = x + 2 + \dfrac{5}{{2x - 1}}\\
P \in Z\\
\Leftrightarrow x + 2 + \dfrac{5}{{2x - 1}} \in Z\\
\Leftrightarrow \dfrac{5}{{2x - 1}} \in Z\left( {x \in Z} \right)\\
\Leftrightarrow \left( {2x - 1} \right) \in U\left( 5 \right) = \left\{ { - 5; - 1;1;5} \right\}\\
\Leftrightarrow x \in \left\{ { - 2;0;1;3} \right\}\\
x \in Z;x < 0 \Rightarrow x = - 2
\end{array}$
Vậy $x = - 2$
$\begin{array}{l}
b)A = {n^3} + 20n = n\left( {{n^2} + 20} \right)\\
n \vdots 2 \Rightarrow n = 2k\left( {k \in Z} \right)\\
\Rightarrow A = 2k\left( {{{\left( {2k} \right)}^2} + 20} \right) = 2k\left( {4{k^2} + 20} \right) = 8k\left( {{k^2} + 5} \right)\\
\Rightarrow A \vdots 8
\end{array}$
Ta có đpcm.
B3:
$\begin{array}{l}
a)\dfrac{{2021 - x}}{{2020}} + \dfrac{{4139 - x}}{{2019}} + \dfrac{{6055 - x}}{{2018}} = 6\\
\Leftrightarrow \left( {\dfrac{{2021}}{{2020}} + \dfrac{{4139}}{{2019}} + \dfrac{{6055}}{{2018}}} \right) - x\left( {\dfrac{1}{{2020}} + \dfrac{1}{{2019}} + \dfrac{1}{{2018}}} \right) = 6\\
\Leftrightarrow \left( {1 + \dfrac{1}{{2020}} + 2 + \dfrac{1}{{2019}} + 3 + \dfrac{1}{{2018}}} \right) - x\left( {\dfrac{1}{{2020}} + \dfrac{1}{{2019}} + \dfrac{1}{{2018}}} \right) = 6\\
\Leftrightarrow \left( {\dfrac{1}{{2020}} + \dfrac{1}{{2019}} + \dfrac{1}{{2018}}} \right) - x\left( {\dfrac{1}{{2020}} + \dfrac{1}{{2019}} + \dfrac{1}{{2018}}} \right) = 0\\
\Leftrightarrow \left( {\dfrac{1}{{2020}} + \dfrac{1}{{2019}} + \dfrac{1}{{2018}}} \right)\left( {1 - x} \right) = 0\\
\Leftrightarrow 1 - x = 0\\
\Leftrightarrow x = 1
\end{array}$
Vậy $x=1$
b) ĐKXĐ: $x\ne 2019$
Ta có:
$\begin{array}{l}
\dfrac{{2020 - x}}{{2019 - x}} > 2\\
\Leftrightarrow \dfrac{{2020 - x}}{{2019 - x}} - 2 > 0\\
\Leftrightarrow \dfrac{{2020 - x - 2\left( {2019 - x} \right)}}{{2019 - x}} > 0\\
\Leftrightarrow \dfrac{{x - 2018}}{{2019 - x}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2018 > 0\\
2019 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2018 < 0\\
2019 - x < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2018\\
x < 2019
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2018\\
x > 2019
\end{array} \right.\left( l \right)
\end{array} \right. \Leftrightarrow 2018 < x < 2019
\end{array}$
Vậy tập nghiệm của bất phương trình là: $S = \left\{ {x \in R|2018 < x < 2019} \right\}$
