Đáp án:
\(\begin{array}{l}
a,\,\,2.\left( {2x - 3y} \right)\\
b,\,\,x.\left( {\dfrac{1}{2}{x^2} - 5xy + 1} \right)\\
c,\,\,xy.\left( {3{x^2} - 6 + 8{x^2}y} \right)\\
d,\,\,\,2\left( {x - y} \right)\left( {y - 2} \right)\\
e,\,\,\,x\left( {x - y} \right)\left( {{x^2} - y} \right)\\
f,\,\,\,3.\left( {y - x} \right)\left( {x + 2y} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\,\,4x - 6y = 2.2x - 2.3y = 2.\left( {2x - 3y} \right)\\
b,\,\,\dfrac{1}{2}{x^3} - 5{x^2}y + x = x.\dfrac{1}{2}{x^2} - x.5xy + x = x.\left( {\dfrac{1}{2}{x^2} - 5xy + 1} \right)\\
c,\,\,3{x^3}y - 6xy + 8{x^3}{y^2}\\
= xy.3{x^2} - xy.6 + xy.8{x^2}y\\
= xy.\left( {3{x^2} - 6 + 8{x^2}y} \right)\\
d,\,\,\,2x\left( {y - 2} \right) - 2y.\left( {y - 2} \right)\\
= \left( {y - 2} \right).\left( {2x - 2y} \right)\\
= \left( {y - 2} \right).2.\left( {x - y} \right)\\
= 2\left( {x - y} \right)\left( {y - 2} \right)\\
e,\,\,\,{x^3}\left( {x - y} \right) - xy\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {{x^3} - xy} \right)\\
= \left( {x - y} \right).x.\left( {{x^2} - y} \right)\\
= x\left( {x - y} \right)\left( {{x^2} - y} \right)\\
f,\,\,\,3x\left( {y - x} \right) + 6y\left( {y - x} \right)\\
= \left( {y - x} \right).\left( {3x + 6y} \right)\\
= \left( {y - x} \right).3.\left( {x + 2y} \right)\\
= 3.\left( {y - x} \right)\left( {x + 2y} \right)
\end{array}\)
