Toán Lớp 8: giúp mình với cần gấp
* Dùng hằng đẳng thức
Bài 1 : phân tích các đa thức sau thành nhân tử
a. x mũ 3 + 6x mũ 2 + 12x + 8
b. x mũ 3 – 3x mũ 2 + 3x -1
c. 1 – 9x + 27x mũ 2 – 27x mũ 3
d. 4b mũ 2 c mũ 2 – (b mũ 2 + c mũ 2 – a mũ 2 ) tất cả mũ 2
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a,\\
{\left( {x + 2} \right)^3}\\
b,\\
{\left( {x – 1} \right)^3}\\
c,\\
{\left( {1 – 3x} \right)^3}\\
d,\\
– \left( {b – c – a} \right).\left( {b – c + a} \right).\left( {b + c – a} \right).\left( {b + c + a} \right)
\end{array}\)
a,\\
{x^3} + 6{x^2} + 12x + 8\\
= {x^3} + 3.2.{x^2} + 3.4.x + 8\\
= {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3}\\
= {\left( {x + 2} \right)^3}\\
b,\\
{x^3} – 3{x^2} + 3x – 1\\
= {x^3} – 3.{x^2}.1 + 3.x{.1^2} – {1^3}\\
= {\left( {x – 1} \right)^3}\\
c,\\
1 – 9x + 27{x^2} – 27{x^3}\\
= 1 – 3.3x + 3.9{x^2} – {3^3}.{x^3}\\
= {1^3} – {3.1^2}.3x + 3.1.{\left( {3x} \right)^2} – {\left( {3x} \right)^3}\\
= {\left( {1 – 3x} \right)^3}\\
d,\\
4{b^2}{c^2} – {\left( {{b^2} + {c^2} – {a^2}} \right)^2}\\
= {\left( {2bc} \right)^2} – {\left( {{b^2} + {c^2} – {a^2}} \right)^2}\\
= \left[ {2bc – \left( {{b^2} + {c^2} – {a^2}} \right)} \right].\left[ {2bc + \left( {{b^2} + {c^2} – {a^2}} \right)} \right]\\
= \left[ {2bc – {b^2} – {c^2} + {a^2}} \right].\left[ {2bc + {b^2} + {c^2} – {a^2}} \right]\\
= – \left[ { – 2bc + {b^2} + {c^2} – {a^2}} \right].\left[ {2bc + {b^2} + {c^2} – {a^2}} \right]\\
= – \left[ {\left( {{b^2} – 2bc + {c^2}} \right) – {a^2}} \right].\left[ {\left( {{b^2} + 2bc + {c^2}} \right) – {a^2}} \right]\\
= – \left[ {{{\left( {b – c} \right)}^2} – {a^2}} \right].\left[ {{{\left( {b + c} \right)}^2} – {a^2}} \right]\\
= – \left[ {\left( {b – c} \right) – a} \right].\left[ {\left( {b – c} \right) + a} \right].\left[ {\left( {b + c} \right) – a} \right].\left[ {\left( {b + c} \right) + a} \right]\\
= – \left( {b – c – a} \right).\left( {b – c + a} \right).\left( {b + c – a} \right).\left( {b + c + a} \right)
\end{array}\)