Đáp án:

\(\begin{array}{l}
4,\\
a,\\
7700\\
b,\\
2500\\
c,\\
8000000\\
5,\\
a,\\
\left[ \begin{array}{l}
x = 0\\
x =  \pm \sqrt {13} 
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 2000\\
x = \dfrac{1}{5}
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d,\\
\left[ \begin{array}{l}
x = 0\\
x =  - \dfrac{1}{5}
\end{array} \right.\\
d,\\
\left[ \begin{array}{l}
x =  - 1\\
x = 0
\end{array} \right.\\
e,\\
x = 0\\
f,\\
\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
g,\\
\left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{5}
\end{array} \right.\\
h,\\
\left[ \begin{array}{l}
x = 3\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
4,\\
a,\\
A = {x^2} + xy + x = x.\left( {x + y + 1} \right)\\
x = 77;y = 22 \Rightarrow A = 77.\left( {77 + 22 + 1} \right) = 77.100 = 7700\\
b,\\
B = x\left( {x - y} \right) + y.\left( {y - x} \right)\\
 = x\left( {x - y} \right) + y.\left[ { - \left( {x - y} \right)} \right]\\
 = x\left( {x - y} \right) - y\left( {x - y} \right)\\
 = \left( {x - y} \right).\left( {x - y} \right)\\
 = {\left( {x - y} \right)^2}\\
x = 53;\,\,y = 3 \Rightarrow B = {\left( {53 - 3} \right)^2} = {50^2} = 2500\\
c,\\
C = x.\left( {x - 1} \right) - y.\left( {1 - x} \right)\\
 = x\left( {x - 1} \right) - y.\left[ { - \left( {x - 1} \right)} \right]\\
 = x\left( {x - 1} \right) + y\left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {x + y} \right)\\
x = 2001;\,\,y = 1999 \Rightarrow C = \left( {2001 - 1} \right).\left( {2001 + 1999} \right) = 2000.4000 = 8000000\\
5,\\
a,\\
{x^3} - 13x = 0\\
 \Leftrightarrow x\left( {{x^2} - 13} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} - 13 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 13
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  \pm \sqrt {13} 
\end{array} \right.\\
b,\\
5x\left( {x - 2000} \right) - x + 2000 = 0\\
 \Leftrightarrow 5x\left( {x - 2000} \right) - \left( {x - 2000} \right) = 0\\
 \Leftrightarrow \left( {x - 2000} \right)\left( {5x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2000 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2000\\
x = \dfrac{1}{5}
\end{array} \right.\\
c,\\
2x\left( {x - 2} \right) + 3\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {2x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
2x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d,\\
x + 5{x^2} = 0\\
 \Leftrightarrow x.\left( {1 + 5x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
1 + 5x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - \dfrac{1}{5}
\end{array} \right.\\
d,\\
x + 1 = {\left( {x + 1} \right)^2}\\
 \Leftrightarrow {\left( {x + 1} \right)^2} - \left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right).\left[ {\left( {x + 1} \right) - 1} \right] = 0\\
 \Leftrightarrow \left( {x + 1} \right).x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = 0
\end{array} \right.\\
e,\\
{x^3} + x = 0\\
 \Leftrightarrow x\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} =  - 1\,\,\,\,\left( {L,\,\,{x^2} \ge 0} \right)
\end{array} \right. \Rightarrow x = 0\\
f,\\
x\left( {x - 2} \right) + x - 2 = 0\\
 \Leftrightarrow x\left( {x - 2} \right) + \left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
g,\\
5x\left( {x - 3} \right) - x + 3 = 0\\
 \Leftrightarrow 5x\left( {x - 3} \right) - \left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {5x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{5}
\end{array} \right.\\
h,\\
{x^2}\left( {x - 3} \right) + 12 - 4x = 0\\
 \Leftrightarrow {x^2}\left( {x - 3} \right) - \left( {4x - 12} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x - 3} \right) - 4.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)