Question

18. O coeficiente de x⁵ no desenvolvimento de
(Vx +³Vx}¹² é igual a:
a) 1
b) 66
c) 220
d) 792
e) 924
obs: O V é raiz quadrada.​

Answer 1

Resposta:

E

Explicação passo-a-passo:

O $(k+1)-$ésimo termo do desenvolvimento de $(\sqrt{x}+\sqrt[3]{x})^{12}$ é:

$\dbinom{12}{k}\cdot(\sqrt{x})^{12-k}\cdot(\sqrt[3]{x})^{k}$

Como queremos o termo de coeficiente $x^5$, devemos ter:

$(\sqrt{x})^{12-k}\cdot(\sqrt[3]{x})^{k}=x^5$

$(x^{\frac{1}{2}})^{12-k}\cdot(x^{\frac{1}{3}})^{k}=x^5$

$x^{\frac{12-k}{2}}\cdot x^{\frac{k}{3}}=x^5$

$x^{\frac{12-k}{2}+\frac{k}{3}}=x^5$

Igualando os expoentes:

$\dfrac{12-k}{2}+\dfrac{k}{3}=5$

$3\cdot(12-k)+2k=6\cdot5$

$36-3k+2k=30$

$3k-2k=36-30$

$k=6$

Assim, o termo procurado é:

$\dbinom{12}{6}\cdot(\sqrt{x})^{6}\cdot(\sqrt[3]{x})^6$

$=\dfrac{12!}{6!\cdot6!}\cdot x^3\cdot x^2$

$=\dfrac{12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6!}{6!\cdot6!}\cdot x^5$

$=\dfrac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1}\cdot x^5$

$=2\cdot11\cdot2\cdot3\cdot7\cdot x^5$

$=924x^5$

Letra E

Answer 2

Resposta:

Alternativa e)

Explicação passo-a-passo:

Binômio de Newton

(A+B)ⁿ

$\displaystyle T_{p+1}={n \choose p}A^{n-p}.B^p$

$\displaystyle (\sqrt{x}+\sqrt[3]{x})^{12}\\\\ A=\sqrt{x} =x^{\frac{1}{2} }\\\\B=\sqrt[3]{x} =x^{\frac{1}{3} }\\\\n=12$

$\displaystyle T_{p+1}={12 \choose p}.x^{\frac{1}{2}.(12-p)}.x^{\frac{1}{3}.p}={12 \choose p}.x^{(\frac{12}{2}-\frac{p}{2}+\frac{p}{3})}={12 \choose p}.x^{(\frac{36-3p+2p}{6})}=\\\\\\{12 \choose p}.x^{(\frac{36-p}{6})}$

Como queremos o valor de x⁵:

$\displaystyle x^{\frac{36-p}{6}}=x^5\\\\{\frac{36-p}{6}}=5\\\\36-p=30\\p=36-30=6$

O termo para que o valor de x⁵ será:

$\displaystyle T_{7}={12 \choose 6}.x^{(\frac{36-6}{6})}={12 \choose 6}.x^{(\frac{30}{6})}={12 \choose 6}.x^5$

O coeficiente do valor de x⁵ será

$\displaystyle {12 \choose 6}=\frac{12!}{(12-6)!6!} =\frac{12.11.10.9.8.7.\diagup\!\!\!\! 6!}{6!\diagup\!\!\!\! 6!} =\frac{12.11.10.9.8.7}{6.5.4.3.2.1} =924$