Đáp án:
$\begin{array}{l}
g)\left( {x - y} \right)\left( {{x^2} - {y^2} + x} \right)\\
h)2\left( {x - 2} \right)\left( {x + 2} \right)\left( {2{x^2} - 17} \right)\\
i)x\left( {x - 4} \right)\left( {x - 11} \right)\\
k)\left( {x + 1} \right)\left( {x - 3} \right)\left( {x + 2} \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
g)x{\left( {x - y} \right)^2} + y{\left( {x - y} \right)^2} - xy + {x^2}\\
= \left( {x - y} \right)\left[ {x\left( {x - y} \right) + y\left( {x - y} \right) + x} \right]\\
= \left( {x - y} \right)\left[ {\left( {x + y} \right)\left( {x - y} \right) + x} \right]\\
= \left( {x - y} \right)\left( {{x^2} - {y^2} + x} \right)\\
h){\left( {8 - 2{x^2}} \right)^2} - 18\left( {x + 2} \right)\left( {x - 2} \right)\\
= {\left( {2\left( {4 - {x^2}} \right)} \right)^2} - 18\left( {{x^2} - 4} \right)\\
= 4{\left( {{x^2} - 4} \right)^2} - 18\left( {{x^2} - 4} \right)\\
= \left( {{x^2} - 4} \right)\left( {4\left( {{x^2} - 4} \right) - 18} \right)\\
= 2\left( {x - 2} \right)\left( {x + 2} \right)\left( {2{x^2} - 17} \right)\\
i){x^3} - 16x - 15x\left( {x - 4} \right)\\
= \left( {{x^3} - 16x} \right) - 15x\left( {x - 4} \right)\\
= x\left( {{x^2} - 16} \right) - 15x\left( {x - 4} \right)\\
= x\left( {x - 4} \right)\left( {x + 4} \right) - 15x\left( {x - 4} \right)\\
= x\left( {x - 4} \right)\left[ {\left( {x + 4} \right) - 15} \right]\\
= x\left( {x - 4} \right)\left( {x - 11} \right)\\
k){x^3} - 7x - 6\\
= {x^3} + {x^2} - {x^2} - x - 6x - 6\\
= \left( {x + 1} \right)\left( {{x^2} - x - 6} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - 3x + 2x - 6} \right)\\
= \left( {x + 1} \right)\left( {x - 3} \right)\left( {x + 2} \right)
\end{array}$
