Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
14,\\
3x\left( {y + 2} \right) - 3\left( {y + 2} \right) = \left( {y + 2} \right)\left( {3x - 3} \right)\\
= 3.\left( {y + 2} \right)\left( {x - 1} \right)\\
15,\\
{x^2} - {y^2} - 2x + 2y = \left( {{x^2} - {y^2}} \right) + \left( { - 2x + 2y} \right)\\
= \left( {x - y} \right)\left( {x + y} \right) - 2.\left( {x - y} \right) = \left( {x - y} \right)\left( {x + y + 2} \right)\\
16,\\
2x + 2y - {x^2} - xy = \left( {2x + 2y} \right) - \left( {{x^2} + xy} \right)\\
= 2.\left( {x + y} \right) - x\left( {x + y} \right) = \left( {x + y} \right)\left( {2 - x} \right)\\
17,\\
{x^2} - 2x - 4{y^2} - 4y = \left( {{x^2} - 4{y^2}} \right) - \left( {2x + 4y} \right)\\
= \left( {x - 2y} \right)\left( {x + 2y} \right) - 2.\left( {x + 2y} \right) = \left( {x + 2y} \right)\left( {x - 2y - 2} \right)\\
19,\\
{x^2}\left( {x - 1} \right) + 16.\left( {1 - x} \right) = {x^2}\left( {x - 1} \right) - 16.\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} - 16} \right) = \left( {x - 1} \right)\left( {x - 4} \right)\left( {x + 4} \right)\\
20,\\
2{x^2} + 3x - 2xy - 3y = \left( {2{x^2} - 2xy} \right) + \left( {3x - 3y} \right)\\
= 2x.\left( {x - y} \right) + 3.\left( {x - y} \right) = \left( {x - y} \right)\left( {2x + 3} \right)\\
21,\\
{x^3} - 4{x^2} + 4x = x.\left( {{x^2} - 4x + 4} \right)\\
= x.\left( {{x^2} - 2.x.2 + {2^2}} \right) = x.{\left( {x - 2} \right)^2}\\
22,\\
15{x^2}y + 20x{y^2} - 25xy\\
= 5xy.3x + 5xy.4y - 5xy.5\\
= 5xy\left( {3x + 4y - 5} \right)
\end{array}\)
Em xem lại đề câu 18 nhé!
