Question

O termo independente de x no desenvolvimento de (x-1/3x)^8 é:

Answer 1

Basta obter x⁰.
$T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right)\ x^k\cdot\big( \frac{1}{3x} \big)^{8-k}\\ \\T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right)\ x^k\cdot\big( \frac{1}{3} \big)^{8-k}\cdot\big( \frac{1}{x} \big)^{8-k}\\ \\T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right) \big( \frac{1}{3} \big)^{8-k}\cdot\big x^k\cdot( \frac{1}{x} \big)^{8-k}$
$\\T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right) \big( \frac{1}{3} \big)^{8-k}\cdot\big x^k\cdot \big( x^{-1} \big)^{8-k}\\ \\T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right) \big( \frac{1}{3} \big)^{8-k}\cdot\big x^k\cdot x^{-8+k}\\ \\T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right) \big( \frac{1}{3} \big)^{8-k}\cdot x^{k+(-8+k)}\\ \\T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right) \big( \frac{1}{3} \big)^{8-k}\cdot x^{k-8+k}$
$T_{k+1}=\left( \begin{array}{c} 8 \\ k \end{array} \right) \big( \frac{1}{3} \big)^{8-k}\cdot x^{2k-8}$
Para que tenhamos x⁰ teremos que fazer 2k - 8 = 0
2k - 8 = 0 ⇒ 2k = 0 + 8 ⇒ 2k = 8 ⇒ k = 8/2⇒ k = 4.
Logo, se k = 0 teremos x⁰.  
$T_{4+1}=\left( \begin{array}{c} 8 \\ 4 \end{array} \right) \big( \frac{1}{3} \big)^{8-4}\cdot x^{2\cdot4-8}= \frac{8!}{(8-4)!4!}\cdot \big( \frac{1}{3} \big)^4\cdot x^0=\frac{8\cdot7\cdot6\cdot5\cdot4!}{4!\cdot4\cdot3\cdot2\cdot1!}\cdot \frac{1}{81}\cdot x^0\\ T_5=70\cdot \frac{1}{81}\cdot x^0\\ \\T_5= \frac{70}{81}\cdot x^0\\ \\T_5= \frac{70}{81}$