phân tích các đa thức sau thành nhân tử
a)1+2xy-x^2-y^2
b) a^2+b^2-c^2-d^2-2ab+2cd
c)a^3b^3-1
d)x^2(y-z)+y^2(z-x)+z^2(x-y)
e)x^2-15x+36
f)x^12-3x^6y^6+2y^12
g)x^8-64x^2
h)(x^2-8)^2-784
phân tích các đa thức sau thành nhân tử a)1+2xy-x^2-y^2 b) a^2+b^2-c^2-d^2-2ab+2cd c)a^3b^3-1 d)x^2(y-z)+y^2(z-x)+z^2(x-y) e)x^2-15x+36 f)x^12-3x^6y^6
By Julia
Đáp án:
Giải thích các bước giải: a) 1+2xy-x^2-y^2=1-(x^2-2xy+y^2)
=1-(x-y)^2
=(1-x+y).(1+x-y)
\(\begin{array}{l}
a)\,\,\,1 + 2xy – {x^2} – {y^2} = 1 – \left( {{x^2} – 2xy + {y^2}} \right)\\
= 1 – {\left( {x – y} \right)^2} = \left( {1 – x + y} \right)\left( {1 + x – y} \right).\\
b)\,\,{a^2} + {b^2} – {c^2} – {d^2} – 2ab + 2cd\\
= \left( {{a^2} – 2ab + {b^2}} \right) – \left( {{c^2} – 2cd + {d^2}} \right)\\
= {\left( {a – b} \right)^2} – {\left( {c – d} \right)^2}\\
= \left( {a – b – c + d} \right)\left( {a – b + c – d} \right).\\
c)\,\,{a^3}{b^3} – 1 = \left( {ab – 1} \right)\left[ {{{\left( {ab} \right)}^2} – ab + 1} \right].\\
d)\,\,{x^2}\left( {y – z} \right) + {y^2}\left( {z – x} \right) + {z^2}\left( {x – y} \right)\\
= {x^2}\left( {y – z} \right) + {y^2}\left( {z – y + y – x} \right) + {z^2}\left( {x – y} \right)\\
= {x^2}\left( {y – z} \right) + {y^2}\left( {z – y} \right) + {y^2}\left( {y – x} \right) + {z^2}\left( {x – y} \right)\\
= {x^2}\left( {y – z} \right) – {y^2}\left( {y – z} \right) + {y^2}\left( {y – x} \right) – {z^2}\left( {y – x} \right)\\
= \left( {y – z} \right)\left( {{x^2} – {y^2}} \right) – \left( {x – y} \right)\left( {{y^2} – {z^2}} \right)\\
= \left( {y – z} \right)\left( {x – y} \right)\left( {x + y} \right) – \left( {x – y} \right)\left( {y – z} \right)\left( {y + z} \right)\\
= \left( {y – z} \right)\left( {x – y} \right)\left( {x + y – y – z} \right)\\
= \left( {y – z} \right)\left( {x – y} \right)\left( {x – z} \right).\\
e)\,\,{x^2} – 15x + 36\\
= {x^2} – 3x – 12x + 36\\
= x\left( {x – 3} \right) – 12\left( {x – 3} \right)\\
= \left( {x – 3} \right)\left( {x – 12} \right).\\
f)\,\,\,{x^{12}} – 3{x^6}{y^6} + 2{y^{12}}\\
= {x^{12}} – {x^6}{y^6} – 2{x^6}{y^6} + 2{y^{12}}\\
= {x^6}\left( {{x^6} – {y^6}} \right) – 2{y^6}\left( {{x^6} – {y^6}} \right)\\
= \left( {{x^6} – {y^6}} \right)\left( {{x^6} – 2{y^6}} \right)\\
= \left( {{x^3} + {y^3}} \right)\left( {{x^3} – {y^3}} \right)\left( {{x^6} – 2{y^6}} \right)\\
= \left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right)\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {{x^6} – 2{y^6}} \right).\\
g)\,\,{x^8} – 64{x^2} = {x^2}\left( {{x^4} – 64} \right)\\
= {x^2}\left( {{x^2} – 8} \right)\left( {{x^2} + 8} \right).\\
h)\,\,\,{\left( {{x^2} – 8} \right)^2} – 784 = \left( {{x^2} – 8 – 28} \right)\left( {{x^2} – 8 + 28} \right)\\
= \left( {{x^2} – 36} \right)\,\left( {{x^2} + 20} \right) = \left( {x – 6} \right)\left( {x + 6} \right)\left( {{x^2} + 20} \right).
\end{array}\)