Giải thích các bước giải:

a.Ta có:

$$\begin{align}(x+3)^7&=\sum _{i=0}^7\binom{7}{i}x^{\left(7-i\right)}\cdot \:3^i\\&=\frac{7!}{0!\left(7-0\right)!}x^7\cdot \:3^0+\frac{7!}{1!\left(7-1\right)!}x^6\cdot \:3^1+\frac{7!}{2!\left(7-2\right)!}x^5\cdot \:3^2+\frac{7!}{3!\left(7-3\right)!}x^4\cdot \:3^3+\frac{7!}{4!\left(7-4\right)!}x^3\cdot \:3^4+\frac{7!}{5!\left(7-5\right)!}x^2\cdot \:3^5+\frac{7!}{6!\left(7-6\right)!}x^1\cdot \:3^6+\frac{7!}{7!\left(7-7\right)!}x^0\cdot \:3^7\\&=x^7+21x^6+189x^5+945x^4+2835x^3+5103x^2+5103x+2187\end{align}$$

b.Ta có:

$$\begin{align}(x-2y)^4&=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\left(-2y\right)^i\\ &=\frac{4!}{0!\left(4-0\right)!}x^4\left(-2y\right)^0+\frac{4!}{1!\left(4-1\right)!}x^3\left(-2y\right)^1+\frac{4!}{2!\left(4-2\right)!}x^2\left(-2y\right)^2+\frac{4!}{3!\left(4-3\right)!}x^1\left(-2y\right)^3+\frac{4!}{4!\left(4-4\right)!}x^0\left(-2y\right)^4\\&=x^4-8x^3y+24x^2y^2-32xy^3+16y^4  \end{align}$$