Cho dãy số $(X_n)$ $\left\{\begin{matrix}X_1=1 \\ X_{n+1}=\sqrt{X^2_n+X_n+1}-\sqrt{X^2_n-X_n+1}\end{matrix}\right.$ 
a) CMR: dãy số trên có giới hạn 
b) Tìm giới hạn đó
chả nhớ gì về cái này cả –  banhquykeomut 19-11-12 09:06 PM
khó nhờ.. –  gaara.sshn 19-11-12 08:57 PM
uh. Lâu rồi t ko làm dạng này –  kellyhoang297 19-11-12 08:33 PM
đề hay quá –  luffykunneu 19-11-12 08:26 PM
a)
Ta có: $X_n>0,\forall n\in\mathbb{N}^*$.
Thật vậy: $X_1>0$.
Với $X_k>0\Rightarrow X_k^2+X_k+1>X_k^2-X_k+1$
$\Rightarrow X_{k+1}>0$
Theo quy nạp suy ra: $X_n>0,\forall n\in\mathbb{N}^*$.
Lại có:
     $\sqrt{X_n^2+X_n+1}+\sqrt{X_n^2-X_n+1}$
$=\sqrt{\left(X_n+\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2}+\sqrt{\left(\frac{1}{2}-X_n\right)^2+\left(\frac{\sqrt3}{2}\right)^2}$
$\ge\sqrt{\left(X_n+\frac{1}{2}+\frac{1}{2}-X_n\right)^2+\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)^2}=2$
Suy ra:
$X_{n+1}=\frac{2X_n}{\sqrt{X_n^2+X_n+1}+\sqrt{X_n^2-X_n+1}}$
           $\le\frac{2X_n}{2}=X_n$
Suy ra: $X_n$ là dãy giảm và bị chặn dưới $\Rightarrow \exists \lim X_n$.

b) Giả sử: $\lim X_n=L$
Suy ra: $L=\sqrt{L^2+L+1}-\sqrt{L^2-L+1}\Leftrightarrow L=0$
Vậy: $\lim X_n=0$
đáp án hay –  banhquykeomut 19-11-12 09:06 PM
vote vote –  phong8 19-11-12 09:06 PM
ngon rồi/// –  gaara.sshn 19-11-12 08:57 PM
đáp án chuẩn luôn –  kellyhoang297 19-11-12 08:33 PM
thanks nha –  luffykunneu 19-11-12 08:27 PM
Ta có: $X_{n+1}=\frac{2X_n}{\sqrt{X_n^2+X_n+1}+\sqrt{X_n^2-X_n+1}} (*)$
Áp dụng BĐT $\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}$ với $a=c=\frac{\sqrt{3}}{2},b=\frac{1}{2}+X_n,d=\frac{1}{2}-X_n$ ta được: $\sqrt{X_n^2+X_n+1}+\sqrt{X_n^2-X_n+1}\geq 2.$
Do đó $X_{n+1}\leq X_n,\forall n$ hay dãy $(X_n)$ là dãy giảm.
Vì $X_n>0,\forall n$ nên tồn tại $\lim_{n\to \infty}X_n=L\geq 0$.
Từ $(*)$ cho $n\to \infty$ ta được: $L=\frac{2L}{\sqrt{L^2+L+1}+\sqrt{L^2-L+1}}$ suy ra $L=0$.
vote vote –  phong8 19-11-12 09:06 PM
hay hay hay –  gaara.sshn 19-11-12 08:58 PM
thanks nha –  luffykunneu 19-11-12 08:27 PM

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