1)chứng minh rằng với mọi x, y ta luôn có: $x^2 + y^2+1\geq xy+x+y$
2) với n là số tự nhiên . Tính
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{1+3}}+\frac{1}{\sqrt{1}+\sqrt{1+3}+\sqrt{1+3+5}}+...+\frac{1}{\sqrt{1}+\sqrt{1+3}+\sqrt{1+3+5}+...+\sqrt{1+3+5+...+(2n+1)}}$
bài 1:
$x^2+y^2+1\geq xy+x+y\Leftrightarrow 2(x^2+y^2+1)\geq 2(xy+x+y)\Leftrightarrow (x-y)^2+(x-1)^2+(y-1)^2\geq 0$  
(luôn đúng)
=>đpcm
B1:
$\Leftrightarrow (x-y)^2+(x-1)^2+(y-1)^2=\geq 0(LĐ)$
$2.\;$ Ta có các tính chất sau (chứng minh bằng quy nạp)
$1+3+5+\ldots+(2n+1)=n^2$
$1+2+3+\ldots+n=\frac{n(n+1)}{2}\Rightarrow \frac 1{1+2+3+\ldots+n}=\frac 2n-\frac 2{n+1}$

$P=\frac{1}{\sqrt{1^2}}+\frac{1}{\sqrt{1^2}+\sqrt{2^2}}+\frac{1}{\sqrt{1^2}+\sqrt{2^2}+\sqrt{3^2}}+\ldots\frac{1}{\sqrt{1^2}+\sqrt{2^2}+\ldots+\sqrt{n^2}}$
$\quad=\frac 11+\frac 1{1+2}+\frac 1{1+2+3}+\ldots+\frac 1{1+2+3+\ldots+n}$
$\quad=\frac 21-\frac 22+\frac 22-\frac 23+\frac 23-\frac 24+\ldots+\frac 2{n}-\frac 2{n+1}$
$\quad=\boxed{\frac{2n}{n+1}}$

uh mình nhầm chỗ đó, nhưng bên dưới vẫn đúng nhé –  tran85295 21-12-17 08:59 PM
bạn ơi bài trên hình như sai r.1 3 5 ... (2n 1)=(n 1)^2 chứ nhỉ:v –  puu 21-12-17 08:29 PM

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