Chủ đề này có 2 trả lời

[DOTOANNANG]

$\lceil$ https://diendantoanh...11/#entry717818 $\rfloor$

Giả sử rằng:

$\text{V}_{\,1}\left ( x_{\,1},\,x_{\,2},\,x_{\,3} \right )= \text{N}^{\,2}$ , khi đó:

$$\text{V}_{\,k}\left ( x_{\,1},\,x_{\,2},\,x_{\,3} \right )\leqq \left [ 1+ k\,\sum\limits_{\,m= 0\,}^{\left [ \frac{k}{2} \right ]}\,\frac{\left ( -\,1 \right )^{\,m}\left ( k- m- 1 \right )\text{!}}{m\text{!}\left ( k- 2\,m \right )\text{!}}\left ( -\,1+ 2\text{N} \right )^{\,k- 2\,m} \right ]^{\,2}$$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 05-12-2018 - 21:00

[DOTOANNANG]

$\lceil$ https://diendantoanh...11/#entry717818 $\rfloor$

Với $r\geqq 1$ thì:

$$\sqrt{r+ \sqrt{\text{V}_{\,2}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )- n^{\,2}}}\leqq r+ \sqrt{\text{V}_{\,1}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )- n^{\,2}}$$

[DOTOANNANG]

Với $r\geqq 1$ thì:

$$\sqrt{r+ \sqrt{\text{V}_{\,2}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )- n^{\,2}}}\leqq r+ \sqrt{\text{V}_{\,1}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )- n^{\,2}}$$

$$\sqrt{r+ \sqrt{\text{V}_{\,\mathit{p\,0}}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )- n^{\,2}}}\leqq r+ \sqrt{\text{V}_{\,\mathit{p\,1}}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )- n^{\,2}}$$ với $0\leqq \mathit{p0}\leqq \mathit{p1}\leqq 1$