Chủ đề này có 26 trả lời

[Viettk14nbk]

$sqrt{a*b*c}$

[Viettk14nbk]

$U_{n}^2 > U_{n-1}*U_{n+1}$, $\forall n \ge 1$



$\lim_{\frac{1}{n^2}\left ( \frac{1}{U_{1}}+\frac{2}{U_{2}}+...+\frac{n}{U_{n}} \right )}$

[Viettk14nbk]

U_{n}^2 > U_{n-1}*U_{n+1}$, $\forall n \ge 1



\lim_{\frac{1}{n^2}\left ( \frac{1}{U_{1}}+\frac{2}{U_{2}}+...+\frac{n}{U_{n}} \right )}

[Viettk14nbk]

$U_{n}^2 > U_{n-1}.U_{n+1}$, $\forall n \ge 1$



$\lim_{x \rightarrow \infty} \frac{1}{n^2}\left ( \frac{1}{U_{1}}+\frac{2}{U_{2}}+...+\frac{n}{U_{n} \right )}$

[Viettk14nbk]

$U_{n}^2 > U_{n-1}.U_{n+1}$, $\forall n \ge 1$



$\lim_{x \to \infty } \frac{1}{n^2}\left ( \frac{1}{U_{1}}+\frac{2}{U_{2}}+...+\frac{n}{U_{n} \right )}$

[Viettk14nbk]

$U_{n}^2 > U_{n-1}.U_{n+1}$, $\forall n \ge 1$



$\lim_{x \to \infty } \frac{1}{n^2}\left ( \frac{1}{U_{1}}+\frac{2}{U_{2}}+...+\frac{n}{U_{n}} \right )$

[Viettk14nbk]

$\left(x_{n}\right)$

[Viettk14nbk]

$x_{1}=1; x_{n+1}=\frag{n+1}{n+2}x_{n}+n^2$

[Viettk14nbk]

$x_{1}=1; x_{n+1}=\fraq{n+1}{n+2}x_{n}+n^2$

[Viettk14nbk]

$x_{1}=1; x_{n+1}=\frac{n+1}{n+2}x_{n}+n^2$

[Viettk14nbk]

$\lim_{n \rightarrow 0} \frac{\sqrt[3]{x_{n}}}}{n+1}$

[Viettk14nbk]

$\lim_{n \to \infty}\frac{\sqrt[3]{x_{n}}}{n+1}$

[Viettk14nbk]

$\lim_{n \to \infty}\left(\frac{\sqrt[3]{x_{n}}}{n+1}\right)$

[Viettk14nbk]

$n=1,2,...$

[Viettk14nbk]

$x_{n+1}=\frac{1}{16}(4x_{n}+1+4\sqrt{4x_{n}+1})$ $(1\leq n\in \mathbb{N})$

[Viettk14nbk]

$\sqrt{£€¥|^§……|¥£[%=\%[{<><§!,}$

[mathstu]

thử ké

564

nguon

124

Bài viết đã được chỉnh sửa nội dung bởi mathstu: 31-05-2016 - 21:57

[mathstu]

$\boxed{\text{Bài toán 1.}}$

$ABC$

$A,B,C$

Bài viết đã được chỉnh sửa nội dung bởi mathstu: 01-06-2016 - 00:00

[mathstu]

$$\left\{\begin{matrix} e^x=y+\sqrt{z^2+1}\\ e^y=z+\sqrt{x^2+1}\\ e^z=x+\sqrt{y^2+1} \end{matrix}\right.$$

$x^{2}$

 $\measuredangle A <90$

 

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Bài viết đã được chỉnh sửa nội dung bởi mathstu: 01-06-2016 - 00:49

[anhmattroi97]

$\sqrt{x+1}$