Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
5x\left( {x - y} \right) + 3y - 3x\\
 = 5x.\left( {x - y} \right) - \left( {3x - 3y} \right)\\
 = 5x.\left( {x - y} \right) - 3.\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {5x - 3} \right)\\
b,\\
{x^3} - 4{x^2}y + 4x{y^2} - 25x\\
 = x.\left[ {\left( {{x^2} - 4xy + 4{y^2}} \right) - 25} \right]\\
 = x.\left[ {{{\left( {x - 2y} \right)}^2} - {5^2}} \right]\\
 = x.\left( {x - 2y - 5} \right)\left( {x - 2y + 5} \right)\\
c,\\
{x^2} - 5x + 4 = \left( {{x^2} - x} \right) - \left( {4x - 4} \right)\\
 = x\left( {x - 1} \right) - 4\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 4} \right)\\
d,\\
{\left( {2x - 1} \right)^2} - 2.\left( {2x - 1} \right).\left( {x + 2} \right) + {\left( {x + 2} \right)^2}\\
 = {\left[ {\left( {2x - 1} \right) - \left( {x + 2} \right)} \right]^2}\\
 = {\left( {x - 3} \right)^2}\\
2,\\
a,\\
{103^2} = {\left( {100 + 3} \right)^2} = {100^2} + 2.100.3 + {3^2} = 10000 + 600 + 9 = 10609\\
b,\\
95.105 = \left( {100 - 5} \right)\left( {100 + 5} \right) = {100^2} - {5^2} = 10000 - 25 = 9975\\
c,\\
{195^2} = {\left( {200 - 5} \right)^2} = {200^2} - 2.200.5 + {5^2} = 40000 - 2000 + 25 = 38025\\
d,\\
{75^2} + 75.50 + {25^2} = {75^2} + 2.75.25 + {25^2} = {\left( {75 + 25} \right)^2} = {100^2} = 10000\\
3,\\
a,\\
3x\left( {x - 2} \right) + 1 = 3{x^2} - 8\\
 \Leftrightarrow 3{x^2} - 6x + 1 = 3{x^2} - 8\\
 \Leftrightarrow  - 6x =  - 9\\
 \Leftrightarrow x = \frac{3}{2}\\
b,\\
\left( {x - 3} \right)\left( {x + 1} \right) - \left( {x + 2} \right)\left( {x - 2} \right) = 5\\
 \Leftrightarrow \left( {{x^2} - 2x - 3} \right) - \left( {{x^2} - 4} \right) = 5\\
 \Leftrightarrow {x^2} - 2x - 3 - {x^2} + 4 = 5\\
 \Leftrightarrow  - 2x = 4\\
 \Leftrightarrow x =  - 2\\
c,\\
\left( {3x - 1} \right)\left( {x + 2} \right) - 2x - 4 = 0\\
 \Leftrightarrow \left( {3x - 1} \right)\left( {x + 2} \right) - \left( {2x + 4} \right) = 0\\
 \Leftrightarrow \left( {3x - 1} \right)\left( {x + 2} \right) - 2.\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left( {x + 2} \right).\left[ {\left( {3x - 1} \right) - 2} \right] = 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {3x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
3x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = 1
\end{array} \right.\\
d,\\
{x^2} - 5x + 6 = 0\\
 \Leftrightarrow \left( {{x^2} - 2x} \right) - \left( {3x - 6} \right) = 0\\
 \Leftrightarrow x\left( {x - 2} \right) - 3.\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x - 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.
\end{array}\)