Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
13{x^3}{y^2}{z^2} - 18{x^2}{y^2}{z^4} = {z^2}{y^2}{z^2}\left( {13x - 18{z^2}} \right)\\
b,\\
{x^2} - 2xy + 3x - 3y + {y^2}\\
= \left( {{x^2} - 2xy + {y^2}} \right) + \left( {3x - 3y} \right)\\
= {\left( {x - y} \right)^2} + 3.\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x - y + 3} \right)\\
c,\\
5x{y^2} - 10xyz + 5x{z^2}\\
= 5x\left( {{y^2} - 2yz + {z^2}} \right)\\
= 5x{\left( {y - z} \right)^2}\\
d,\\
2{x^2} - 9x + 10\\
= \left( {2{x^2} - 4x} \right) - \left( {5x - 10} \right)\\
= 2x\left( {x - 2} \right) - 5.\left( {x - 2} \right)\\
= \left( {x - 2} \right)\left( {2x - 5} \right)\\
e,\\
{x^3} - 3{x^2} - 4x + 12\\
= \left( {{x^3} - 3{x^2}} \right) - \left( {4x - 12} \right)\\
= {x^2}\left( {x - 3} \right) - 4.\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
f,\\
27{x^3} + 27{x^2} + 9x + 1\\
= {\left( {3x} \right)^3} + 3.{\left( {3x} \right)^2}.1 + 3.3x{.1^2} + {1^3}\\
= {\left( {3x + 1} \right)^3}\\
g,\\
{x^2} - 5x + 4\\
= \left( {{x^2} - x} \right) - \left( {4x - 4} \right)\\
= x\left( {x - 1} \right) - 4.\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {x - 4} \right)\\
h,\\
5{x^2} + 5{y^2} - 10xy - 125{z^2}\\
= 5.\left[ {\left( {{x^2} - 2xy + {y^2}} \right) - 25{z^2}} \right]\\
= 5.\left[ {{{\left( {x - y} \right)}^2} - {{\left( {5z} \right)}^2}} \right]\\
= 5.\left( {x - y - 5z} \right)\left( {x - y + 5z} \right)\\
j,\\
6{x^2} - 11x + 3\\
= \left( {6{x^2} - 2x} \right) - \left( {9x - 3} \right)\\
= 2x.\left( {3x - 1} \right) - 3.\left( {3x - 1} \right)\\
= \left( {3x - 1} \right)\left( {2x - 3} \right)\\
k,\\
2{x^2} - 5xy + 3{y^2}\\
= \left( {2{x^2} - 2xy} \right) - \left( {3xy - 3{y^2}} \right)\\
= 2x.\left( {x - y} \right) - 3y\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {2x - 3y} \right)
\end{array}\)
