Đáp án:
$\begin{array}{l}
a){x^2} - 10x + 9\\
= {x^2} - x - 9x + 9\\
= \left( {x - 1} \right)\left( {x - 9} \right)\\
b)2{x^2} - 5x + 2\\
= 2{x^2} - 4x - x + 2\\
= \left( {x - 2} \right)\left( {2x - 1} \right)\\
c)3{x^2} - 10xy + 3{y^2}\\
= 3{x^2} - 9xy - xy + 3{y^2}\\
= \left( {x - 3y} \right)\left( {3x - y} \right)\\
d)2xy - {x^2} + 3{y^2} - 4y + 1\\
= 4{y^2} - 4y + 1 - \left( {{x^2} - 2xy + {y^2}} \right)\\
= {\left( {2y - 1} \right)^2} - {\left( {x - y} \right)^2}\\
= \left( {2y - 1 - x + y} \right)\left( {2x - 1 + x - y} \right)\\
= \left( {x + y - 1} \right)\left( {3x - y - 1} \right)\\
g)4{x^{16}} + 81\\
= {\left( {2{x^8}} \right)^2} + 2.2{x^8}.9 + {9^2} - 36{x^8}\\
= {\left( {2{x^8} + 9} \right)^2} - \left( {36{x^8}} \right)\\
= \left( {2{x^8} + 9 + 6{x^4}} \right)\left( {2{x^8} + 9 - 6{x^4}} \right)\\
i){\left( {5 - y} \right)^6} - 2\left( {125 - 75y + 15{y^2} - {y^3}} \right) + 1\\
= {\left( {5 - y} \right)^6} - 2.{\left( {5 - y} \right)^3} + 1\\
= {\left( {{{\left( {5 - y} \right)}^3} - 1} \right)^2}\\
= {\left( {4 - y} \right)^2}.{\left[ {{{\left( {5 - y} \right)}^2} + 5 - y + 1} \right]^2}\\
= {\left( {4 - y} \right)^2}.{\left( {{y^2} - 11y + 31} \right)^2}\\
k){x^4} + 2018{x^2} + 2017x + 2018\\
= {x^4} - x + 2018{x^2} + 2018x + 2018\\
= x\left( {{x^3} - 1} \right) + 2018\left( {{x^2} + x + 1} \right)\\
= x\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + 2018\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 2018} \right)
\end{array}$
