\[\begin{array}{l}
a)\,\,6{x^2} - 11x + 3 = 6{x^2} - 2x - 9x + 3\\
= 2x\left( {3x - 1} \right) - 3\left( {3x - 1} \right) = \left( {3x - 1} \right)\left( {2x - 3} \right).\\
b)\,\,2{x^2} + 3x - 27 = 2{x^2} - 6x + 9x - 27\\
= 2x\left( {x - 3} \right) + 9\left( {x - 3} \right) = \left( {x - 3} \right)\left( {2x + 9} \right).\\
c)\,\,2{x^2} - 5xy - 3{y^2} = 2{x^2} + xy - 6xy - 3{y^2}\\
= x\left( {2x + y} \right) - 3y\left( {2x + y} \right) = \left( {2x + y} \right) = \left( {x - 3y} \right).\\
d)\,\,4{x^4} + 1 = {\left( {2{x^2}} \right)^2} + 2{x^2} + 1 - 2{x^2} = {\left( {2{x^2} + 1} \right)^2} - 2{x^2}\\
= \left( {2{x^2} + 1 - \sqrt 2 x} \right)\left( {2{x^2} + 1 + \sqrt 2 x} \right).\\
e)\,\,4{x^4} + {y^4} = {\left( {2{x^2}} \right)^2} + 2.2{x^2}.{y^2} + {\left( {{y^2}} \right)^2} - 4{x^2}{y^2}\\
= {\left( {2{x^2} + {y^2}} \right)^2} - {\left( {2xy} \right)^2} = \left( {2{x^2} + {y^2} - 2xy} \right)\left( {2{x^2} + {y^2} + 2xy} \right).
\end{array}\]
Những câu dưới em làm tương tự nhé!!!
