(ITA-04) O termo independente de \mathbf{x} no desenvolvimento do binỏmio \left(\sqrt{\frac{3 \sqrt[3]{x}}{5 x}}-\sqrt[3]{\frac{5 x}{3 \sqrt{x}}}\right)^{12} é:
a) 729 \sqrt[3]{45}
b) 972 \sqrt[3]{15}
c) 891 \sqrt[3]{\frac{3}{5}}
d) 376 \sqrt[3]{\frac{5}{3}}
e) 165 \sqrt[3]{75}
Solução: Alternativa E
O termo geral de um binômio (a+b)^{n} é dado por T_{k+1}=\binom{n}{k} a^{n-k} b^{k} T_{k+1}=\binom{12}{k}\left(\sqrt{\frac{3 \sqrt[3]{x}}{5 x}}\right)^{12-k}\left(-\sqrt[3]{\frac{5 x}{3 \sqrt{x}}}\right)^{k} \Rightarrow T_{k+1}=\binom{12}{k}\left(\sqrt{\frac{3}{5}}\right)^{12-k}\left(-\sqrt[3]{\frac{5}{3}}\right)^{k}\left(x^{11 / 6-1 / 2}\right)^{12-k}\left(x^{1 / 3-1 / 6}\right)^{k} \Rightarrow T_{k+1}=\binom{12}{k}\left(\sqrt{\frac{3}{5}}\right)^{12-k}\left(-\sqrt[3]{\frac{5}{3}}\right)^{k} x^{-\frac{1}{3}(12-k)+\frac{k}{6}}
Para obtermos o termo independente, temos que ter: -\frac{1}{3}(12-k)+\frac{1}{6} \cdot k=0 \Rightarrow k=8 Substituindo k=8, temos : \mathrm{T}_{9}=\binom{12}{8}\left(\sqrt{\frac{3}{5}}\right)^{12-8}\left(-\sqrt[3]{\frac{5}{3}}\right)^{8} \Rightarrow \mathrm{~T}_{9}=\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2} \cdot \frac{9}{25} \cdot \frac{25}{9} \cdot\left(\sqrt[3]{\frac{5}{3}}\right)^{2} \Rightarrow \mathrm{T}_{y}=495 \cdot \sqrt[3]{\frac{5^{2}}{3^{2}}} \Rightarrow \mathrm{~T}_{y}=495 \cdot \sqrt[3]{\frac{5^{2}}{3^{2}}} \cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}} \Rightarrow \mathrm{~T}_{y}=495 \cdot \frac{\sqrt[3]{75}}{3} \Rightarrow \mathrm{~T}_{y}=165 \sqrt[3]{75}