Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} + 5x + 5y - {y^2}\\
= \left( {{x^2} - {y^2}} \right) + \left( {5x + 5y} \right)\\
= \left( {x - y} \right)\left( {x + y} \right) + 5\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x - y + 5} \right)\\
b,\\
2{x^3} - 8x = 2x\left( {{x^2} - 4} \right) = 2x\left( {x - 2} \right)\left( {x + 2} \right)\\
c,\\
{x^2}y - 2x{y^2} + {y^3} - 9y\\
= y\left( {{x^2} - 2xy + {y^2} - 9} \right)\\
= y.\left[ {\left( {{x^2} - 2xy + {y^2}} \right) - 9} \right]\\
= y.\left[ {{{\left( {x - y} \right)}^2} - {3^2}} \right]\\
= y.\left( {x - y - 3} \right)\left( {x - y + 3} \right)\\
d,\\
2xy + 16 - {x^2} - {y^2}\\
= 16 - \left( {{x^2} - 2xy + {y^2}} \right)\\
= {4^2} - {\left( {x - y} \right)^2}\\
= \left( {4 - x + y} \right)\left( {4 + x - y} \right)\\
e,\\
2{x^2} + 4x + 2 - 2{y^2} = 2.\left[ {\left( {{x^2} + 2x + 1} \right) - {y^2}} \right]\\
= 2.\left[ {{{\left( {x + 1} \right)}^2} - {y^2}} \right] = 2.\left( {x + 1 - y} \right)\left( {x + 1 + y} \right)\\
f,\\
{x^2} + 5x - 3xy - 15y = x\left( {x + 5} \right) - 3y\left( {x + 5} \right) = \left( {x + 5} \right)\left( {x - 3y} \right)\\
g,\\
{x^3} - 2{x^2} + x - x{y^2}\\
= x.\left( {{x^2} - 2x + 1 - {y^2}} \right)\\
= x.\left[ {{{\left( {x - 1} \right)}^2} - {y^2}} \right]\\
= x.\left( {x - 1 - y} \right)\left( {x - 1 + y} \right)\\
h,\\
{x^2} - 4xy + 6x - 12y + 4{y^2}\\
= \left( {{x^2} - 4xy + 4{y^2}} \right) + \left( {6x - 12y} \right)\\
= {\left( {x - 2y} \right)^2} + 6.\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {x - 2y + 6} \right)\\
k,\\
{x^2} - 4x - 4y - {y^2}\\
= \left( {{x^2} - {y^2}} \right) - \left( {4x + 4y} \right)\\
= \left( {x - y} \right)\left( {x + y} \right) - 4.\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x - y - 4} \right)
\end{array}\)
