Question

clasa a 10 a
Rezolvarea va rog

Answer 1

$T_{k+1}=C_n^ka^{n-k}b^k\\a=\sqrt[3]x=x^\frac13\\b=-x\sqrtx=-x\cdotx^\frac12=-x^\frac32\\T_{k+1}=C_{11}^k\left(x^\frac13\right)^{11-k}\left(-x^\frac32\right)^k=C_{11}^kx^{\frac{11-k}{3}}(-1)^kx^{\frac{3k}{2}}=\\=(-1)^kC_{11}^kx^{\frac{11-k}3+\frac{3k}2}=(-1)^kC_{11}^k x^\frac{22+7k}{6}$

$a)\:\displaystyle T_4=T_{3+1}=(-1)^3C_{11}^3x^\frac{22+7\cdot3}6=-\frac{11!}{3!\cdot 8!}x^\frac{43}6=\\=-\frac{8!\cdot9\cdot10\cdot11}{2\cdot3\cdot 8!}x^\frac{43}6=-165x^\frac{43}6\\\\T_9=T_{8+1}=(-1)^8C_{11}^8x^\frac{22+7\cdot8}6=\frac{11!}{3!\cdot 8!}x^\frac{78}6=\\=\frac{8!\cdot9\cdot10\cdot11}{2\cdot3\cdot 8!}x^{13}=165x^{13}$

$b)\: x^{\frac{22+7k}{6}}=x^{-1}\Rightarrow\displaystyle \frac{22+7k}6=-1\Rightarrow 22+7k=-6\\\Rightarrow 7k=-28\Rightarrow k=-4<0\\\Rightarrow \text{nu exist\u a termenul care \^il con\c tine pe }x^{-1}$

$c)\:x^\frac{22+7k}6=x^6\Rightarrow\displaystyle\frac{22+7k}6=6\Rightarrow 22+7k=36\Rightarrow 7k=14\Rightarrow k=2\\\text{Termenul care \^il con\c tine pe }x^6 \text{ este }T_{2+1}=T_3$