展开 $$$\left(x - 2\right)^{5}$$$

展开 $$$\left(x - 2\right)^{5}$$$。

该展开由以下公式给出：$$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$，其中$$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$和$$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$。

我们有$$$a = x$$$、$$$b = -2$$$和$$$n = 5$$$。

因此，$$$\left(x - 2\right)^{5} = \sum_{k=0}^{5} {\binom{5}{k}} x^{5 - k} \left(-2\right)^{k}$$$。

现在，计算 $$$k$$$ 从 $$$0$$$ 到 $$$5$$$ 的每个取值对应的乘积。

$$$k = 0$$$: $$${\binom{5}{0}} x^{5 - 0} \left(-2\right)^{0} = \frac{5!}{\left(5 - 0\right)! 0!} x^{5 - 0} \left(-2\right)^{0} = x^{5}$$$

$$$k = 1$$$: $$${\binom{5}{1}} x^{5 - 1} \left(-2\right)^{1} = \frac{5!}{\left(5 - 1\right)! 1!} x^{5 - 1} \left(-2\right)^{1} = - 10 x^{4}$$$

$$$k = 2$$$: $$${\binom{5}{2}} x^{5 - 2} \left(-2\right)^{2} = \frac{5!}{\left(5 - 2\right)! 2!} x^{5 - 2} \left(-2\right)^{2} = 40 x^{3}$$$

$$$k = 3$$$: $$${\binom{5}{3}} x^{5 - 3} \left(-2\right)^{3} = \frac{5!}{\left(5 - 3\right)! 3!} x^{5 - 3} \left(-2\right)^{3} = - 80 x^{2}$$$

$$$k = 4$$$: $$${\binom{5}{4}} x^{5 - 4} \left(-2\right)^{4} = \frac{5!}{\left(5 - 4\right)! 4!} x^{5 - 4} \left(-2\right)^{4} = 80 x$$$

$$$k = 5$$$: $$${\binom{5}{5}} x^{5 - 5} \left(-2\right)^{5} = \frac{5!}{\left(5 - 5\right)! 5!} x^{5 - 5} \left(-2\right)^{5} = -32$$$

因此，$$$\left(x - 2\right)^{5} = x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32$$$。

$$$\left(x - 2\right)^{5} = x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32$$$A