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A melhor resposta!
2014-07-12T21:48:04-03:00
Olá Cristy,

dado o binômio,

(3a-4)^4

podemos usar a fórmula binomial,

 T_{p+1} =\left(\begin{array}{ccc}n\\p\end{array}\right)*a^{n-p}*b~^p

e desenvolvê-lo, acompanhe:

(3a-4)^4=  \left(\begin{array}{ccc}4\\0\end{array}\right)*3a^{4-0}*(-4)^0\\\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\to~\dfrac{4!}{0!(4-0)!}= \dfrac{4!}{1!4!}=1*3a^4*1  \\\\
(3a-4)^4=\left(\begin{array}{ccc}4\\1\end{array}\right)*3a^{4-1}*(-4)^1\\\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\to~\dfrac{4!}{1!(4-1)!}= \dfrac{4!}{1!3!}=4*3a^3*(-4)


(3a-4)^4=  \left(\begin{array}{ccc}4\\2\end{array}\right)*3a^{4-2} *(-4)^2\\\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \dfrac{4!}{2!(4-2)!}= \dfrac{4!}{2!2!}=6*3a^2*16\\\\\\
(3a-4)^4=  \left(\begin{array}{ccc}4\\3\end{array}\right)*3a^{4-3}*(-4)^3\\\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \dfrac{4!}{3!(4-3)!}= \dfrac{4!}{3!1!}=6*3a^1*(-64)


(3a-4)^4=  \left(\begin{array}{ccc}4\\4\end{array}\right)*3a^{4-4}*(-4)^4\\\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \dfrac{4!}{4!(4-4)!}= \dfrac{4!}{4!0!}=1*3a^0*256\\\\\\
(3a-4)^4=1*3a^4*1+4*3a^3*(-4)+6*3a^2*16+6*3a^1*(-64)\\
+1*3a^0*256\\\\O~desenvolvimento~do~binomio~sera:\\\\
\boxed{(3a-4)^4=3a^4-48a^3+288a^2-1.152a+256}

Espero ter ajudado e tenha ótimos estudos =))
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