Đáp án:
$\left(x^3-\dfrac{2}{x^3}\right)^8=-\displaystyle\sum_{k=0}^8C^k_8.x^{6k-24}.2^{8-k}\\ \left(2x^4+\dfrac{1}{x^3}\right)^5=\displaystyle\sum_{k=0}^8C^k_8.2^k.x^{7k-15}$
Giải thích các bước giải:
$\left(x^3-\dfrac{2}{x^3}\right)^8\\ =\displaystyle\sum_{k=0}^8C^k_8.(x^3)^k.\left(-\dfrac{2}{x^3}\right)^{8-k}\\ =-\displaystyle\sum_{k=0}^8C^k_8.x^{3k}.\dfrac{2^{8-k}}{x^{3(8-k)}}\\ =-\displaystyle\sum_{k=0}^8C^k_8.x^{3k-3(8-k)}.2^{8-k}\\ =-\displaystyle\sum_{k=0}^8C^k_8.x^{6k-24}.2^{8-k}\\ \left(2x^4+\dfrac{1}{x^3}\right)^5\\ =\displaystyle\sum_{k=0}^5C^k_5.(2x^4)^k.\left(\dfrac{1}{x^3}\right)^{5-k}\\ =\displaystyle\sum_{k=0}^5C^k_5.2^k.x^{4k}.\dfrac{1}{x^{3(5-k)}}\\ =\displaystyle\sum_{k=0}^5C^k_5.2^k.x^{4k-3(5-k)}\\ =\displaystyle\sum_{k=0}^5C^k_5.2^k.x^{7k-15}$
