Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
{a^4} + a{b^3} - {a^3}b - {b^4}\\
= \left( {{a^4} + a{b^3}} \right) - \left( {{a^3}b + {b^4}} \right)\\
= a\left( {{a^3} + {b^3}} \right) - b\left( {{a^3} + {b^3}} \right)\\
= \left( {a - b} \right)\left( {{a^3} + {b^3}} \right)\\
= \left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\
c,\\
9{x^2} - 24xy + 16{y^2}\\
= {\left( {3x} \right)^2} - 2.3x.4y + {\left( {4y} \right)^2}\\
= {\left( {3x - 4y} \right)^2}\\
d,\\
{x^2} + 2xy - {z^2} - 2zt + {y^2} - {t^2}\\
= \left( {{x^2} + 2xy + {y^2}} \right) - \left( {{z^2} + 2zt + {t^2}} \right)\\
= {\left( {x + y} \right)^2} - {\left( {z + t} \right)^2}\\
= \left[ {\left( {x + y} \right) - \left( {z + t} \right)} \right].\left[ {\left( {x + y} \right) + \left( {z + t} \right)} \right]\\
= \left( {x + y - z - t} \right)\left( {x + y + z + t} \right)\\
e,\\
12y - 9{x^2} - 36 - 3{x^2}y\\
= \left( {12y + 36} \right) - \left( {9{x^2} + 3{x^2}y} \right)\\
= 12\left( {y + 3} \right) - 3{x^2}\left( {3 + y} \right)\\
= \left( {3 + y} \right)\left( {12 - 3{x^2}} \right)\\
= \left( {3 + y} \right).3.\left( {4 - {x^2}} \right)\\
= 3.\left( {y + 3} \right).\left( {2 - x} \right).\left( {2 + x} \right)\\
f,\\
4{x^3} + 16x = 4x.\left( {{x^2} + 4} \right)
\end{array}\)
Em xem lại đề câu a nhé?
