Đáp án:

h. \(\left( {x - y + 4} \right)\left( {x - y - 1} \right)\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left( {x - 2} \right)\left( {x + 2} \right) + {\left( {x - 2} \right)^2}\\
 = \left( {x - 2} \right)\left( {x + 2 + x - 2} \right)\\
 = 2x\left( {x - 2} \right)\\
b.x\left( {{x^2} - 2x + 1 - {y^2}} \right)\\
 = x\left[ {{{\left( {x - 1} \right)}^2} - {y^2}} \right]\\
 = x\left[ {\left( {x - 1 - y} \right)\left( {x - 1 + y} \right)} \right]\\
 = x\left( {x - y - 1} \right)\left( {x + y - 1} \right)\\
c.{x^3} - 4{x^2} - 12x + 27\\
 = {x^3} + 3{x^2} - 7{x^2} - 21x + 9x + 27\\
 = {x^2}\left( {x + 3} \right) - 7x\left( {x + 3} \right) + 9\left( {x + 3} \right)\\
 = \left( {x + 3} \right)\left( {{x^2} - 7x + 9} \right)\\
d.{\left( {6x + 1} \right)^2} + 2\left( {6x + 1} \right)\left( {6x - 1} \right) + {\left( {6x - 1} \right)^2}\\
 = {\left( {6x + 1 + 6x - 1} \right)^2}\\
 = {\left( {12x} \right)^2} = 144{x^2}\\
e.{x^3} + 2{x^2} - 5{x^2} - 10x + 6x + 12\\
 = {x^2}\left( {x + 2} \right) - 5x\left( {x + 2} \right) + 6\left( {x + 2} \right)\\
 = \left( {x + 2} \right)\left( {{x^2} - 5x + 6} \right)\\
 = \left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 3} \right)\\
g.{x^4} - 5{x^2} + 4 = {x^4} - 4{x^2} - {x^2} + 4\\
 = {x^2}\left( {{x^2} - 4} \right) - \left( {{x^2} - 4} \right)\\
 = \left( {{x^2} - 4} \right)\left( {{x^2} - 1} \right)\\
 = \left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right)\\
h.{\left( {x - y} \right)^2} + 3\left( {x - y} \right) - 4\\
 = {\left( {x - y} \right)^2} - \left( {x - y} \right) + 4\left( {x - y} \right) - 4\\
 = \left( {x - y} \right)\left( {x - y - 1} \right) + 4\left( {x - y - 1} \right)\\
 = \left( {x - y + 4} \right)\left( {x - y - 1} \right)
\end{array}\)